Question

Question 1. Dscribe the change you observed when you added 1 mL of 0.1 M KSCN to the 2 mL portion of the diluted solution. Copy the equation from the procedure and explain your observations in terms of LeChatelier’s Principle. Choose the best answer. The additional thiocyantate ion shifted the equilibrium toward the products, producing morehexathiocyanatoferrate(III) lightening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the reactants, producing lesshexathiocyanatoferrate(III) and deepening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the reactants, producing morehexathiocyanatoferrate(III) lightening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the products, producing morehexathiocyanatoferrate(III) and deepening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Question 2. Show your calculations to determine the final volume of the 0.001 M FeCl3 solution. Choose the best answer. M1V1 = M2V2. (0.001 M)V1=(0.1 M)(0.05 L) V1=0.050 L = 50 mL M1V1 = M2V2. (0.001 M)V1=(0.1 M)(0.005 L) V1=0.500 L = 500 mL M1V1 = M2V2. (0.1 M)(0.005 L)=(0.001 M)V2 V2=0.500 L = 500 mL M1V1 = M2V2. (0.1 M)(0.0005 L)=(0.001 M)V2 V2=0.050 L = 50 mL Question 3. Describe the change you observed when you added 1 mL of 0.1 M FeCl3 to the diluted mixture of 0.001 M FeCl3(aq) and 0.1 M KSCN(aq). Refer to the equation in step 1 and explain your observations in terms of LeChatelier’s Principle. Choose the best answer. Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored hexathiocyanatoferrate(III) complex and the intensity of color increased.color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored hexathiocyanatoferrate(III) chloride precipitate. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)6Cl3(s). Additional iron(III) ion caused the equilibrium to shift to the reactants to compensate for the additional added iron(III), which increased the intensity of the red color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Additional iron(III) ion increased the concentration of the red-colored iron(III) ion. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Question 4. Predict and then describe what happened when you added HCl(aq) to the copper nitrate solution. Write down the equation for the reaction. Choose the best answer. The added hydrogen ion shifted the equilibrium toward the products so that more CuCl42-(aq) was createdand the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + 4H2O(l). The added chloride ion shifted the equilibrium toward the products so that more CuCl42-(aq) was created and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + 4H2O(l). The added chloride ion caused the precipitation of CuCl42-(s), which shifted the equilibrium toward the products and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(s) + 4H2O(l). The added hydrogen ion caused the precipitation of CuCl42-(s), which shifted the equilibrium toward the products and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(s) + 4H2O(l). Question 5. Describe the change, and explain what occurred when you added the water (in terms of LeChatelier’s Principle). Choose the best answer. The added water increased the volume and diluted the concentration of all the dissolved ions, which caused the reaction to shift toward the reactants. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). The added water increased the volume and diluted the concentration of all the dissolved ions, which caused the reaction to shift toward the products. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). Adding water increased the water concentration, which caused the reaction to shift toward the products. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). Adding water increased the water concentration, which caused the reaction to shift toward the reactants. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). Question 6. Describe the change, copy the equation, and explain what occurred when you added the sulfuric acid to the potassium chromate solution. Choose the best answer. The additional hydronium ions caused a shift toward the reactant side, decreasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). The additional hydronium ions caused a shift toward the product side, increasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). The additional hydronium ions caused a shift toward the product side, decreasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). The additional hydronium ions caused a shift toward the reactant side, increasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). Question 7. Describe the observed change when you added sodium hydroxide to the solution? Neither Na+ nor OH- ions appear in the chromate-dichromate equilibrium. Why, then, did the equilibrium shift? Choose the best answer. The solution changed to orange. The added hydroxide ions caused the change. The solution changed to yellow. The added sodium ions caused the change. The solution changed to orange. The added sodium ions caused the change. The solution changed to yellow. The added hydroxide ions caused the change. There was no change since neither the Na+(aq) ion nor the OH-(aq) ion appears in the chromate-dichromate equilibrium. Question 8. Explain the color changes in terms of Le Chatelier’s principle. Choose the best answer. Additional hydroxide ion reacts with the hydronium ion (creating water). The additional water on the product side caused the equilibrium to shift to the reactants. The added hydroxide ions decreased the hydronium ion concentration, which shifted the reaction toward the products. The added hydroxide ions decreased the hydronium ion concentration, which shifted the reaction toward the reactants. There was no shift in the equilibrium since neither the Na+(aq) ion nor the OH-(aq) ion appears in the chromate-dichromate equilibrium. Additional hydroxide ion reacts with the hydronium ion (creating water). The additional water on the product side caused the equilibrium to shift to the products. Question 9. Describe what you observed when concentrated hyrochloric acid was added to saturated sodium chloride? Write the equation for the equilibrium, and explain what you observed in terms of LeChatelier’s Principle. Choose the best answer. Excess chloride ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate NaCl was formed (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid increased the acidity of the solution and the precipitate NaCl was formed (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid increased the acidity of the solution and the precipitate NaCl dissolved (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate HCl was formed (see equation). HCl(s) ⇔ H+(aq) + Cl-(aq) Excess chloride ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate NaCl dissolved (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate HCl dissolved (see equation). HCl(s) ⇔ H+(aq) + Cl-(aq) Question 10. Describe the effect water had on the mixture. Explain the change properly in terms of the LeChatelier principle. Choose the best answer. Additonal water increased the volume and reduced the concentrations of sodium ion and chloride ions, which shifted the reaction toward reactants. Additional water caused the sodium chloride precipitate to fall to the bottom of the test tube. Additonal water increased the volume and reduced the concentrations of sodium and chloride ions, which shifted the reaction toward products. Additional water caused an increase in the amount of NaCl(s) in solution, which shifted the equilibrium to the right, dissolving the solid sodium chloride. Question 11. The equilibrium equation shows that SbCl3 reacts with water to form insoluble SbOCl. Why does the solution of antimony(III) chloride have no visible precipitate in it? Choose the best answer. There is very little H+(aq) present. The SbCl3(aq) and HCl(aq) have not yet been added together, so no SbOCl(s) has been created. A substantial amount of HCl(aq) already exists in the solution, so the equilibrium has shifted completely to the reactants and all of the SbOCl(s) has dissolved. The solid SbOCl is present in such a small concentration that it is not visible. Question 12. Describe what happened when you added water to the solution. Copy the equation, and explain your observation. Choose the best answer. The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the greatest moles of solute and solid SbOCl is formed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the greatest moles of solute and solid SbOCl is consumed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the fewest moles of solute and solid SbOCl is formed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the fewest moles of solute and solid SbOCl is consumed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). Question 13. Describe what effect the concentrated hydrochloric acid had? Why? Choose the best answer. Adding HCl shifts the equilibrium toward the reactants, which causes solid SbOCl to dissolve. Adding HCl shifts the equilibrium toward the products, which causes solid SbOCl to form. Adding HCl increases the concentration of water (a reactant), which causes solid SbOCl to dissolve. Adding HCl increases the concentration of water (a reactant), which causes solid SbOCl to form. Question 14. Write the net-ionic equation for the reaction you observed between barium chloride and potassium chromate. Indicate the states of all reactants and products. Choose the best answer. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). Ba2+(aq) + CrO4-(aq) ⇔ Ba(CrO4)2(s). BaCl2(aq) + K2CrO4(aq) ⇔ BaCrO4(s) + K2Cl2(aq). BaCl2(aq) + K2CrO4(aq) ⇔ BaCrO4(s) + 2KCl(aq). Question 15. Describe what happened when HCl(aq) was added? Record both changes that you observed. To explain these observations, first tell what effect the addition of HCl(aq) had on the chromate-dichromate equilibrium (see Step 6). Then explain how that equilibrium shift in turn affected the barium chromate equilibrium (see Step 14). Write both equations again here as part of your explanation. Choose the best answer. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the right decreasing the chromate concentration. This in turn, caused the first equation to shift to the right and the solid barium chromate to form. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the left increasing the chromate concentration. This in turn, caused the first equation to shift to the right and the solid barium chromate to form. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the right decreasing the chromate concentration. This in turn, caused the first equation to shift to the left and the solid barium chromate to dissolve. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the left increasing the chromate concentration. This in turn, caused the first equation to shift to the left and the solid barium chromate to dissolve. Application of Principles. Question 1. The silver ion, Ag+, forms the colorless diamminesilver(I) complex ion, Ag(NH3)2+, that is soluble, when it is in an ammonia solution. If ammonia is added to a solution that contains a AgCl precipitate, the solid dissolves completely. Write a net-ionic equation for the equilibrium involved. Choose the best answer. AgCl(s) + NH3(aq) ⇔ Ag(NH3)2+(aq) + Cl-(aq). Ag+(aq) + 2NH3(aq) ⇔ Ag(NH3)2+(aq). AgCl(s) + 2NH3(aq) ⇔ Ag(NH3)2+(aq). AgCl(s) + 2NH3(aq) ⇔ Ag(NH3)2+(aq) + Cl-(aq). Application of Principles. Question 2. When concentrated H2SO4 (18M) is added to a saturated solution of sodium sulfate, a white precipitate is formed. Write a net-ionic equation for the equilibrium in saturated sodium sulfate, and explain the change in terms of Le Chatelier’s principle. Choose the best answer. 2Na+(aq) + SO42-(aq) + 2H+(aq) + SO42+(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. 2Na+(aq) + SO42-(aq) +H2SO4(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. 2Na+(aq) + SO42-(aq) ⇔ Na2SO4(s). Excess SO42- (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. Na+(aq) + SO4-(aq) ⇔ NaSO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. Na+(aq) + SO4-(aq) ⇔ NaSO4(s). Excess SO42- (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. 2Na+(aq) + SO42-(aq) +H2SO4(aq) ⇔ Na2SO4(s). Excess SO42- (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. 2Na+(aq) + SO42-(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. 2Na+(aq) + SO42-(aq) + 2H+(aq) + SO42+(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. Application of Principles. Question 3. A solution contains the blood-red hexathiocyanatoferrate(III) complex ion, Fe(SCN)63-, in equilibrium with the iron(III) ion, Fe3+, and six thiocyanate ions, SCN-. If the solution is diluted with water, will the color deepen, fade, or stay the same? Why? In addition to predicting whether the solution will deepen, fade, or stay the same, give TWO reasons for that prediction. One of those reasons will involve Le Chatelier’s Principle, and one will not. (At least that will be the case if you make the right prediction!) Choose the best answer. The color of the solution will fade. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with hexathiocyanatoferrate(III) complex ions. 2) The color will also fade because of dilution. The color of the solution will deepen. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with hexathiocyanatoferrate(III) complex ions. 2) The color will also fade because of dilution. The color of the solution will deepen. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with iron(III) and thiocyante ions. 2) The color will also fade because of dilution. The color of the solution will fade. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with iron(III) and thiocyante ions. 2) The color will also fade because of dilution. The color will stay the same. 1) The reactions does not shift to either side. 2) The concentration of the hexathiocyanatoferrate(II) complex ion does not change.

Name: question 1 dscribe the change you observed when you added 1 ml of 01 m kscn to the 2 ml portion of the diluted solution copy the equation from the procedure and explain your observations in 20714
Uploaded: 2023-02-26T16:15:53-08:00
Duration: 2 min 27 s
Channel: Sri K
Description: question 1 dscribe the change you observed when you added 1 ml of 01 m kscn to the 2 ml portion of the diluted solution copy the equation from the procedure and explain your observations in 20714

          Question 1. Dscribe the change you observed
when you added 1 mL of 0.1 M KSCN to the 2 mL portion of the
diluted solution. Copy the equation from the procedure and explain
your observations in terms of LeChatelier’s Principle.
Choose the best answer.
The additional thiocyantate ion shifted the equilibrium toward
the products, producing
morehexathiocyanatoferrate(III) lightening
the color.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).
The additional thiocyantate ion shifted the equilibrium toward
the reactants, producing
lesshexathiocyanatoferrate(III) and
deepening the color.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).
The additional thiocyantate ion shifted the equilibrium toward
the reactants, producing
morehexathiocyanatoferrate(III) lightening
the color.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).
The additional thiocyantate ion shifted the equilibrium toward
the products, producing
morehexathiocyanatoferrate(III) and
deepening the color.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).

Question 2. Show your calculations to determine
the final volume of the 0.001 M FeCl3 solution.
Choose the best answer.
M1V1 =
M2V2.
(0.001 M)V1=(0.1 M)(0.05 L)
V1=0.050 L = 50 mL
M1V1 =
M2V2.
(0.001 M)V1=(0.1 M)(0.005 L)
V1=0.500 L = 500 mL
M1V1 =
M2V2.
(0.1 M)(0.005 L)=(0.001 M)V2
V2=0.500 L = 500 mL
M1V1 =
M2V2.
(0.1 M)(0.0005 L)=(0.001 M)V2
V2=0.050 L = 50 mL

Question 3. Describe the change you observed
when you added 1 mL of 0.1 M FeCl3 to the diluted
mixture of 0.001 M FeCl3(aq) and 0.1 M KSCN(aq). Refer
to the equation in step 1 and explain your observations in terms of
LeChatelier’s Principle. Choose the best
answer.
Additional iron(III) ion shifted the equilibrium toward the
products to form the red-colored hexathiocyanatoferrate(III)
complex and the intensity of color increased.color.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).
Additional iron(III) ion shifted the equilibrium toward the
products to form the red-colored hexathiocyanatoferrate(III)
chloride precipitate.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)6Cl3(s).
Additional iron(III) ion caused the equilibrium to shift to the
reactants to compensate for the additional added iron(III), which
increased the intensity of the red color.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).
Additional iron(III) ion increased the concentration of the
red-colored iron(III) ion.
Fe3+(aq) + 6SCN-(aq) ⇔
Fe(SCN)63-(aq).

Question 4. Predict and then describe what
happened when you added HCl(aq) to the copper nitrate solution.
Write down the equation for the reaction. Choose the best
answer.
The added hydrogen ion shifted the equilibrium
toward the products so that more
CuCl42-(aq) was createdand
the solution changed from blue to green.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(aq) +
4H2O(l).
The added chloride ion shifted the equilibrium
toward the products so that more
CuCl42-(aq) was created and
the solution changed from blue to green.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(aq) +
4H2O(l).
The added chloride ion caused the
precipitation of
CuCl42-(s), which shifted
the equilibrium toward the products and the solution changed from
blue to green.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(s) + 4H2O(l).
The added hydrogen ion caused the
precipitation of
CuCl42-(s), which shifted
the equilibrium toward the products and the solution changed from
blue to green.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(s) + 4H2O(l).

Question 5. Describe the change, and explain
what occurred when you added the water (in terms of LeChatelier’s
Principle). Choose the best answer.
The added water increased the volume and
diluted the concentration of all the dissolved ions, which caused
the reaction to shift toward the reactants.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(aq) + H2O(l).
The added water increased the volume and
diluted the concentration of all the dissolved ions, which caused
the reaction to shift toward the products.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(aq) + H2O(l).
Adding water increased the water concentration,
which caused the reaction to shift toward the
products.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(aq) + H2O(l).
Adding water increased the water concentration,
which caused the reaction to shift toward the
reactants.
Cu(H2O)42+(aq) +
4Cl-(aq) ⇔
CuCl42-(aq) + H2O(l).

Question 6. Describe the change, copy the
equation, and explain what occurred when you added the sulfuric
acid to the potassium chromate solution. Choose the best
answer.
The additional hydronium ions caused a shift toward the
reactant side, decreasing the
concentration of the dichromate ion and turning the solution to an
orange color.
2CrO42-(aq) +
2H3O+(aq) ⇔
Cr2O72-(aq) +
3H2O(l).
The additional hydronium ions caused a shift toward the
product side, increasing the
concentration of the dichromate ion and turning the solution to an
orange color.
2CrO42-(aq) +
2H3O+(aq) ⇔
Cr2O72-(aq) +
3H2O(l).
The additional hydronium ions caused a shift toward the
product side, decreasing the
concentration of the dichromate ion and turning the solution to an
orange color.
2CrO42-(aq) +
2H3O+(aq) ⇔
Cr2O72-(aq) +
3H2O(l).
The additional hydronium ions caused a shift toward the
reactant side, increasing the
concentration of the dichromate ion and turning the solution to an
orange color.
2CrO42-(aq) +
2H3O+(aq) ⇔
Cr2O72-(aq) +
3H2O(l).

Question 7. Describe the observed change when
you added sodium hydroxide to the solution? Neither Na+
nor OH- ions appear in the chromate-dichromate
equilibrium. Why, then, did the equilibrium shift? Choose
the best answer.
The solution changed to orange. The added
hydroxide ions caused the change.
The solution changed to yellow. The added
sodium ions caused the change.
The solution changed to orange. The added
sodium ions caused the change.
The solution changed to yellow. The added
hydroxide ions caused the change.
There was no change since neither the
Na+(aq) ion nor the OH-(aq)
ion appears in the chromate-dichromate equilibrium.

Question 8. Explain the color changes in terms
of Le Chatelier’s principle. Choose the best
answer.
Additional hydroxide ion reacts with the hydronium ion (creating
water). The additional water on the product side caused the
equilibrium to shift to the reactants.
The added hydroxide ions decreased the hydronium ion
concentration, which shifted the reaction toward the
products.
The added hydroxide ions decreased the hydronium ion
concentration, which shifted the reaction toward the
reactants.
There was no shift in the equilibrium since
neither the Na+(aq) ion nor the
OH-(aq) ion appears in the chromate-dichromate
equilibrium.
Additional hydroxide ion reacts with the hydronium ion (creating
water). The additional water on the product side caused the
equilibrium to shift to the products.

Question 9. Describe what you observed when
concentrated hyrochloric acid was added to saturated sodium
chloride? Write the equation for the equilibrium, and explain what
you observed in terms of LeChatelier’s Principle. Choose
the best answer.
Excess chloride ion from the hydrochloric acid shifted the
reaction toward reactants, and the precipitate NaCl was
formed (see equation).
NaCl(s) ⇔ Na+(aq) +
Cl-(aq)
Excess hydrogen ion from the hydrochloric acid increased the
acidity of the solution and the precipitate NaCl was
formed (see equation).
NaCl(s) ⇔ Na+(aq) +
Cl-(aq)
Excess hydrogen ion from the hydrochloric acid increased the
acidity of the solution and the precipitate NaCl
dissolved (see equation).
NaCl(s) ⇔ Na+(aq) +
Cl-(aq)
Excess hydrogen ion from the hydrochloric acid shifted the
reaction toward reactants, and the precipitate HCl was
formed (see equation).
HCl(s) ⇔ H+(aq) +
Cl-(aq)
Excess chloride ion from the hydrochloric acid shifted the
reaction toward reactants, and the precipitate NaCl
dissolved (see equation).
NaCl(s) ⇔ Na+(aq) +
Cl-(aq)
Excess hydrogen ion from the hydrochloric acid shifted the
reaction toward reactants, and the precipitate HCl
dissolved (see equation).
HCl(s) ⇔ H+(aq) +
Cl-(aq)

Question 10. Describe the effect water had on
the mixture. Explain the change properly in terms of the
LeChatelier principle. Choose the best answer.
Additonal water increased the volume and reduced the
concentrations of sodium ion and chloride ions, which shifted the
reaction toward reactants.
Additional water caused the sodium chloride precipitate to
fall to the bottom of the test tube.
Additonal water increased the volume and reduced the
concentrations of sodium and chloride ions, which shifted the
reaction toward products.
Additional water caused an increase in the amount of
NaCl(s) in solution, which shifted the equilibrium to the
right, dissolving the solid sodium chloride.

Question 11. The equilibrium equation shows
that SbCl3 reacts with water to form insoluble SbOCl.
Why does the solution of antimony(III) chloride have no visible
precipitate in it? Choose the best answer.
There is very little H+(aq) present.
The SbCl3(aq) and HCl(aq) have not
yet been added together, so no SbOCl(s) has been
created.
A substantial amount of HCl(aq) already exists in the
solution, so the equilibrium has shifted completely to the
reactants and all of the SbOCl(s) has dissolved.
The solid SbOCl is present in such a small concentration that it
is not visible.

Question 12. Describe what happened when you
added water to the solution. Copy the equation, and explain your
observation. Choose the best answer.
The addition of water increases the volume and dilutes all
dissolved ions. The reaction shifts to the side with the
greatest moles of solute and solid SbOCl is
formed.
SbCl3(aq) + H2O(l) ⇔
SbOCl(s) + 2 HCl(aq).
The addition of water increases the volume and dilutes all
dissolved ions. The reaction shifts to the side with the
greatest moles of solute and solid SbOCl is
consumed.
SbCl3(aq) + H2O(l) ⇔
SbOCl(s) + 2 HCl(aq).
The addition of water increases the volume and dilutes all
dissolved ions. The reaction shifts to the side with the
fewest moles of solute and solid SbOCl is
formed.
SbCl3(aq) + H2O(l) ⇔
SbOCl(s) + 2 HCl(aq).
The addition of water increases the volume and dilutes all
dissolved ions. The reaction shifts to the side with the
fewest moles of solute and solid SbOCl is
consumed.
SbCl3(aq) + H2O(l) ⇔
SbOCl(s) + 2 HCl(aq).

Question 13. Describe what effect the
concentrated hydrochloric acid had? Why? Choose the best
answer.
Adding HCl shifts the equilibrium toward the
reactants, which causes solid SbOCl to
dissolve.
Adding HCl shifts the equilibrium toward the
products, which causes solid SbOCl to
form.
Adding HCl increases the concentration of water (a reactant),
which causes solid SbOCl to dissolve.
Adding HCl increases the concentration of water (a reactant),
which causes solid SbOCl to form.

Question 14. Write the net-ionic equation for
the reaction you observed between barium chloride and potassium
chromate. Indicate the states of all reactants and products.
Choose the best answer.
Ba2+(aq) +
CrO42-(aq) ⇔
BaCrO4(s).
Ba2+(aq) +
CrO4-(aq) ⇔
Ba(CrO4)2(s).
BaCl2(aq) +
K2CrO4(aq) ⇔
BaCrO4(s) +
K2Cl2(aq).
BaCl2(aq) +
K2CrO4(aq) ⇔
BaCrO4(s) + 2KCl(aq).

Question 15. Describe what happened when
HCl(aq) was added? Record both changes that you observed.
To explain these observations, first tell what effect the addition
of HCl(aq) had on the chromate-dichromate equilibrium (see
Step 6). Then explain how that equilibrium shift in turn affected
the barium chromate equilibrium (see Step 14). Write both equations
again here as part of your explanation. Choose the best
answer.
Ba2+(aq) +
CrO42-(aq) ⇔
BaCrO4(s).
2CrO42- + 2H3O+ ⇔
Cr2O72- +
3H2O(l).
Adding HCl shifted the second reaction to the
right decreasing the chromate concentration. This
in turn, caused the first equation to shift to the
right and the solid barium chromate to
form.
Ba2+(aq) +
CrO42-(aq) ⇔
BaCrO4(s).
2CrO42- + 2H3O+ ⇔
Cr2O72- +
3H2O(l).
Adding HCl shifted the second reaction to the left
increasing the chromate concentration. This in turn, caused the
first equation to shift to the right and the solid
barium chromate to form.
Ba2+(aq) +
CrO42-(aq) ⇔
BaCrO4(s).
2CrO42- + 2H3O+ ⇔
Cr2O72- +
3H2O(l).
Adding HCl shifted the second reaction to the
right decreasing the chromate concentration. This
in turn, caused the first equation to shift to the
left and the solid barium chromate to
dissolve.
Ba2+(aq) +
CrO42-(aq) ⇔
BaCrO4(s).
2CrO42- + 2H3O+ ⇔
Cr2O72- +
3H2O(l).
Adding HCl shifted the second reaction to the left
increasing the chromate concentration. This in turn, caused the
first equation to shift to the left and the solid
barium chromate to dissolve.

Application of Principles. Question 1. The
silver ion, Ag+, forms the colorless diamminesilver(I)
complex ion, Ag(NH3)2+, that is
soluble, when it is in an ammonia solution. If ammonia is added to
a solution that contains a AgCl precipitate, the solid dissolves
completely. Write a net-ionic equation for the equilibrium
involved. Choose the best answer.
AgCl(s) + NH3(aq) ⇔
Ag(NH3)2+(aq) +
Cl-(aq).
Ag+(aq) + 2NH3(aq) ⇔
Ag(NH3)2+(aq).
AgCl(s) + 2NH3(aq) ⇔
Ag(NH3)2+(aq).
AgCl(s) + 2NH3(aq) ⇔
Ag(NH3)2+(aq) +
Cl-(aq).

Application of Principles. Question 2. When
concentrated H2SO4 (18M) is added to
a saturated solution of sodium sulfate, a white precipitate is
formed. Write a net-ionic equation for the equilibrium in saturated
sodium sulfate, and explain the change in terms of Le Chatelier’s
principle. Choose the best answer.
2Na+(aq) +
SO42-(aq) +
2H+(aq) +
SO42+(aq) ⇔
Na2SO4(s).
Excess SO42-(aq) (beyond saturation)
from the sulfuric acid causes the reaction to shift to the
right and solid sodium sulfate to
form.
2Na+(aq) +
SO42-(aq)
+H2SO4(aq) ⇔
Na2SO4(s).
Excess SO42-(aq) (beyond saturation)
from the sulfuric acid causes the reaction to shift to the
left and solid sodium sulfate to
dissolve.
2Na+(aq) +
SO42-(aq) ⇔
Na2SO4(s).
Excess SO42- (beyond saturation) from the
sulfuric acid causes the reaction to shift to the
right and solid sodium sulfate to
form.
Na+(aq) +
SO4-(aq) ⇔
NaSO4(s).
Excess SO42-(aq) (beyond saturation)
from the sulfuric acid causes the reaction to shift to the
left and solid sodium sulfate to
dissolve.
Na+(aq) +
SO4-(aq) ⇔
NaSO4(s).
Excess SO42- (beyond saturation) from the
sulfuric acid causes the reaction to shift to the
right and solid sodium sulfate to
form.
2Na+(aq) +
SO42-(aq)
+H2SO4(aq) ⇔
Na2SO4(s).
Excess SO42- (beyond saturation) from the
sulfuric acid causes the reaction to shift to the
right and solid sodium sulfate to
form.
2Na+(aq) +
SO42-(aq) ⇔
Na2SO4(s).
Excess SO42-(aq) (beyond saturation)
from the sulfuric acid causes the reaction to shift to the
left and solid sodium sulfate to
dissolve.
2Na+(aq) +
SO42-(aq) +
2H+(aq) +
SO42+(aq) ⇔
Na2SO4(s).
Excess SO42-(aq) (beyond saturation)
from the sulfuric acid causes the reaction to shift to the
left and solid sodium sulfate to
dissolve.

Application of Principles. Question 3. A
solution contains the blood-red hexathiocyanatoferrate(III) complex
ion, Fe(SCN)63-, in equilibrium with the
iron(III) ion, Fe3+, and six thiocyanate ions,
SCN-. If the solution is diluted with water, will the
color deepen, fade, or stay the same? Why? In addition to
predicting whether the solution will deepen, fade, or stay the
same, give TWO reasons for that prediction. One of those reasons
will involve Le Chatelier’s Principle, and one will not. (At least
that will be the case if you make the right prediction!)
Choose the best answer.
The color of the solution will fade. 1) The
volume will increase, diluting the concentrations of all ions, and
the reaction will shift toward the side with
hexathiocyanatoferrate(III) complex ions. 2) The
color will also fade because of dilution.
The color of the solution will deepen. 1) The
volume will increase, diluting the concentrations of all ions, and
the reaction will shift toward the side with
hexathiocyanatoferrate(III) complex ions. 2) The
color will also fade because of dilution.
The color of the solution will deepen. 1) The
volume will increase, diluting the concentrations of all ions, and
the reaction will shift toward the side with iron(III) and
thiocyante ions. 2) The color will also fade because of
dilution.
The color of the solution will fade. 1) The
volume will increase, diluting the concentrations of all ions, and
the reaction will shift toward the side with iron(III) and
thiocyante ions. 2) The color will also fade because of
dilution.
The color will stay the same. 1) The reactions
does not shift to either side. 2) The concentration of the
hexathiocyanatoferrate(II) complex ion does not change.

Added by Katrina K.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sri K

02/26/2023

Step 1

18 \times 10^{-18} J \times \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\) Given \(n_f = 4\) and \(n_i = 7\), plug in the values: \(\Delta E = -2.18 \times 10^{-18} J \times \left( \frac{1}{4^2} - \frac{1}{7^2} \right)\) \(\Delta E = -2.18 \times 10^{-18} J Show more…

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Question 1. Dscribe the change you observed when you added 1 mL of 0.1 M KSCN to the 2 mL portion of the diluted solution. Copy the equation from the procedure and explain your observations in terms of LeChatelier’s Principle. Choose the best answer. The additional thiocyantate ion shifted the equilibrium toward the products, producing morehexathiocyanatoferrate(III) lightening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the reactants, producing lesshexathiocyanatoferrate(III) and deepening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the reactants, producing morehexathiocyanatoferrate(III) lightening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the products, producing morehexathiocyanatoferrate(III) and deepening the color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Question 2. Show your calculations to determine the final volume of the 0.001 M FeCl3 solution. Choose the best answer. M1V1 = M2V2. (0.001 M)V1=(0.1 M)(0.05 L) V1=0.050 L = 50 mL M1V1 = M2V2. (0.001 M)V1=(0.1 M)(0.005 L) V1=0.500 L = 500 mL M1V1 = M2V2. (0.1 M)(0.005 L)=(0.001 M)V2 V2=0.500 L = 500 mL M1V1 = M2V2. (0.1 M)(0.0005 L)=(0.001 M)V2 V2=0.050 L = 50 mL Question 3. Describe the change you observed when you added 1 mL of 0.1 M FeCl3 to the diluted mixture of 0.001 M FeCl3(aq) and 0.1 M KSCN(aq). Refer to the equation in step 1 and explain your observations in terms of LeChatelier’s Principle. Choose the best answer. Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored hexathiocyanatoferrate(III) complex and the intensity of color increased.color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored hexathiocyanatoferrate(III) chloride precipitate. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)6Cl3(s). Additional iron(III) ion caused the equilibrium to shift to the reactants to compensate for the additional added iron(III), which increased the intensity of the red color. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Additional iron(III) ion increased the concentration of the red-colored iron(III) ion. Fe3+(aq) + 6SCN-(aq) ⇔ Fe(SCN)63-(aq). Question 4. Predict and then describe what happened when you added HCl(aq) to the copper nitrate solution. Write down the equation for the reaction. Choose the best answer. The added hydrogen ion shifted the equilibrium toward the products so that more CuCl42-(aq) was createdand the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + 4H2O(l). The added chloride ion shifted the equilibrium toward the products so that more CuCl42-(aq) was created and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + 4H2O(l). The added chloride ion caused the precipitation of CuCl42-(s), which shifted the equilibrium toward the products and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(s) + 4H2O(l). The added hydrogen ion caused the precipitation of CuCl42-(s), which shifted the equilibrium toward the products and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(s) + 4H2O(l). Question 5. Describe the change, and explain what occurred when you added the water (in terms of LeChatelier’s Principle). Choose the best answer. The added water increased the volume and diluted the concentration of all the dissolved ions, which caused the reaction to shift toward the reactants. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). The added water increased the volume and diluted the concentration of all the dissolved ions, which caused the reaction to shift toward the products. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). Adding water increased the water concentration, which caused the reaction to shift toward the products. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). Adding water increased the water concentration, which caused the reaction to shift toward the reactants. Cu(H2O)42+(aq) + 4Cl-(aq) ⇔ CuCl42-(aq) + H2O(l). Question 6. Describe the change, copy the equation, and explain what occurred when you added the sulfuric acid to the potassium chromate solution. Choose the best answer. The additional hydronium ions caused a shift toward the reactant side, decreasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). The additional hydronium ions caused a shift toward the product side, increasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). The additional hydronium ions caused a shift toward the product side, decreasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). The additional hydronium ions caused a shift toward the reactant side, increasing the concentration of the dichromate ion and turning the solution to an orange color. 2CrO42-(aq) + 2H3O+(aq) ⇔ Cr2O72-(aq) + 3H2O(l). Question 7. Describe the observed change when you added sodium hydroxide to the solution? Neither Na+ nor OH- ions appear in the chromate-dichromate equilibrium. Why, then, did the equilibrium shift? Choose the best answer. The solution changed to orange. The added hydroxide ions caused the change. The solution changed to yellow. The added sodium ions caused the change. The solution changed to orange. The added sodium ions caused the change. The solution changed to yellow. The added hydroxide ions caused the change. There was no change since neither the Na+(aq) ion nor the OH-(aq) ion appears in the chromate-dichromate equilibrium. Question 8. Explain the color changes in terms of Le Chatelier’s principle. Choose the best answer. Additional hydroxide ion reacts with the hydronium ion (creating water). The additional water on the product side caused the equilibrium to shift to the reactants. The added hydroxide ions decreased the hydronium ion concentration, which shifted the reaction toward the products. The added hydroxide ions decreased the hydronium ion concentration, which shifted the reaction toward the reactants. There was no shift in the equilibrium since neither the Na+(aq) ion nor the OH-(aq) ion appears in the chromate-dichromate equilibrium. Additional hydroxide ion reacts with the hydronium ion (creating water). The additional water on the product side caused the equilibrium to shift to the products. Question 9. Describe what you observed when concentrated hyrochloric acid was added to saturated sodium chloride? Write the equation for the equilibrium, and explain what you observed in terms of LeChatelier’s Principle. Choose the best answer. Excess chloride ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate NaCl was formed (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid increased the acidity of the solution and the precipitate NaCl was formed (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid increased the acidity of the solution and the precipitate NaCl dissolved (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate HCl was formed (see equation). HCl(s) ⇔ H+(aq) + Cl-(aq) Excess chloride ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate NaCl dissolved (see equation). NaCl(s) ⇔ Na+(aq) + Cl-(aq) Excess hydrogen ion from the hydrochloric acid shifted the reaction toward reactants, and the precipitate HCl dissolved (see equation). HCl(s) ⇔ H+(aq) + Cl-(aq) Question 10. Describe the effect water had on the mixture. Explain the change properly in terms of the LeChatelier principle. Choose the best answer. Additonal water increased the volume and reduced the concentrations of sodium ion and chloride ions, which shifted the reaction toward reactants. Additional water caused the sodium chloride precipitate to fall to the bottom of the test tube. Additonal water increased the volume and reduced the concentrations of sodium and chloride ions, which shifted the reaction toward products. Additional water caused an increase in the amount of NaCl(s) in solution, which shifted the equilibrium to the right, dissolving the solid sodium chloride. Question 11. The equilibrium equation shows that SbCl3 reacts with water to form insoluble SbOCl. Why does the solution of antimony(III) chloride have no visible precipitate in it? Choose the best answer. There is very little H+(aq) present. The SbCl3(aq) and HCl(aq) have not yet been added together, so no SbOCl(s) has been created. A substantial amount of HCl(aq) already exists in the solution, so the equilibrium has shifted completely to the reactants and all of the SbOCl(s) has dissolved. The solid SbOCl is present in such a small concentration that it is not visible. Question 12. Describe what happened when you added water to the solution. Copy the equation, and explain your observation. Choose the best answer. The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the greatest moles of solute and solid SbOCl is formed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the greatest moles of solute and solid SbOCl is consumed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the fewest moles of solute and solid SbOCl is formed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). The addition of water increases the volume and dilutes all dissolved ions. The reaction shifts to the side with the fewest moles of solute and solid SbOCl is consumed. SbCl3(aq) + H2O(l) ⇔ SbOCl(s) + 2 HCl(aq). Question 13. Describe what effect the concentrated hydrochloric acid had? Why? Choose the best answer. Adding HCl shifts the equilibrium toward the reactants, which causes solid SbOCl to dissolve. Adding HCl shifts the equilibrium toward the products, which causes solid SbOCl to form. Adding HCl increases the concentration of water (a reactant), which causes solid SbOCl to dissolve. Adding HCl increases the concentration of water (a reactant), which causes solid SbOCl to form. Question 14. Write the net-ionic equation for the reaction you observed between barium chloride and potassium chromate. Indicate the states of all reactants and products. Choose the best answer. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). Ba2+(aq) + CrO4-(aq) ⇔ Ba(CrO4)2(s). BaCl2(aq) + K2CrO4(aq) ⇔ BaCrO4(s) + K2Cl2(aq). BaCl2(aq) + K2CrO4(aq) ⇔ BaCrO4(s) + 2KCl(aq). Question 15. Describe what happened when HCl(aq) was added? Record both changes that you observed. To explain these observations, first tell what effect the addition of HCl(aq) had on the chromate-dichromate equilibrium (see Step 6). Then explain how that equilibrium shift in turn affected the barium chromate equilibrium (see Step 14). Write both equations again here as part of your explanation. Choose the best answer. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the right decreasing the chromate concentration. This in turn, caused the first equation to shift to the right and the solid barium chromate to form. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the left increasing the chromate concentration. This in turn, caused the first equation to shift to the right and the solid barium chromate to form. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the right decreasing the chromate concentration. This in turn, caused the first equation to shift to the left and the solid barium chromate to dissolve. Ba2+(aq) + CrO42-(aq) ⇔ BaCrO4(s). 2CrO42- + 2H3O+ ⇔ Cr2O72- + 3H2O(l). Adding HCl shifted the second reaction to the left increasing the chromate concentration. This in turn, caused the first equation to shift to the left and the solid barium chromate to dissolve. Application of Principles. Question 1. The silver ion, Ag+, forms the colorless diamminesilver(I) complex ion, Ag(NH3)2+, that is soluble, when it is in an ammonia solution. If ammonia is added to a solution that contains a AgCl precipitate, the solid dissolves completely. Write a net-ionic equation for the equilibrium involved. Choose the best answer. AgCl(s) + NH3(aq) ⇔ Ag(NH3)2+(aq) + Cl-(aq). Ag+(aq) + 2NH3(aq) ⇔ Ag(NH3)2+(aq). AgCl(s) + 2NH3(aq) ⇔ Ag(NH3)2+(aq). AgCl(s) + 2NH3(aq) ⇔ Ag(NH3)2+(aq) + Cl-(aq). Application of Principles. Question 2. When concentrated H2SO4 (18M) is added to a saturated solution of sodium sulfate, a white precipitate is formed. Write a net-ionic equation for the equilibrium in saturated sodium sulfate, and explain the change in terms of Le Chatelier’s principle. Choose the best answer. 2Na+(aq) + SO42-(aq) + 2H+(aq) + SO42+(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. 2Na+(aq) + SO42-(aq) +H2SO4(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. 2Na+(aq) + SO42-(aq) ⇔ Na2SO4(s). Excess SO42- (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. Na+(aq) + SO4-(aq) ⇔ NaSO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. Na+(aq) + SO4-(aq) ⇔ NaSO4(s). Excess SO42- (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. 2Na+(aq) + SO42-(aq) +H2SO4(aq) ⇔ Na2SO4(s). Excess SO42- (beyond saturation) from the sulfuric acid causes the reaction to shift to the right and solid sodium sulfate to form. 2Na+(aq) + SO42-(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. 2Na+(aq) + SO42-(aq) + 2H+(aq) + SO42+(aq) ⇔ Na2SO4(s). Excess SO42-(aq) (beyond saturation) from the sulfuric acid causes the reaction to shift to the left and solid sodium sulfate to dissolve. Application of Principles. Question 3. A solution contains the blood-red hexathiocyanatoferrate(III) complex ion, Fe(SCN)63-, in equilibrium with the iron(III) ion, Fe3+, and six thiocyanate ions, SCN-. If the solution is diluted with water, will the color deepen, fade, or stay the same? Why? In addition to predicting whether the solution will deepen, fade, or stay the same, give TWO reasons for that prediction. One of those reasons will involve Le Chatelier’s Principle, and one will not. (At least that will be the case if you make the right prediction!) Choose the best answer. The color of the solution will fade. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with hexathiocyanatoferrate(III) complex ions. 2) The color will also fade because of dilution. The color of the solution will deepen. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with hexathiocyanatoferrate(III) complex ions. 2) The color will also fade because of dilution. The color of the solution will deepen. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with iron(III) and thiocyante ions. 2) The color will also fade because of dilution. The color of the solution will fade. 1) The volume will increase, diluting the concentrations of all ions, and the reaction will shift toward the side with iron(III) and thiocyante ions. 2) The color will also fade because of dilution. The color will stay the same. 1) The reactions does not shift to either side. 2) The concentration of the hexathiocyanatoferrate(II) complex ion does not change.