Solubility Product For the following questions, identify the ions comprising the salt, and then use the expression for the solubility product to perform the necessary calculations. This simple exercise involves the assumptions that you can ignore any complications which might result as a consequence of the basicity or acidity of the ions, ion pairing, complex ion formation, or auto-ionization of water. Calculate the concentration, in mol/L, of a saturated aqueous solution of Ca5(PO4)3F (Ksp = 1.00×10-60). 6.11×10-8 mol/L You are correct. Previous Tries Calculate the solubility, in g/100mL, of Ca5(PO4)3F. 3.08×10-6 g/100mL You are correct. Previous Tries Calculate the mass of SrCrO4 (Ksp = 3.60×10-5) which will dissolve in 100 ml of water. 1.22×10-1 g
Added by Robert H.
Close
Step 1
00 x 10⁻⁶⁰. Let the solubility of Ca₅(PO₄)₃F be S mol/L. Then, the concentrations of the ions in the solution are: [Ca²⁺] = 5S [PO₄³⁻] = 3S [F⁻] = S Substitute these values into the Ksp expression: (1.00 x 10⁻⁶⁰) = (5S)⁵(3S)³(S) Now, we need to solve for S: S Show more…
Show all steps
Your feedback will help us improve your experience
Adi S and 55 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Calculating the Molar Solubility The solubility of a substance is the quantity that dissolves to form a saturated solution, and it can be represented by the solubility product expression (Ksp). The solubility product expression is equivalent to the product of the ion concentrations after each is raised to a power that is equal to its coefficient in the balanced equilibrium equation. For an example, see the following reaction involving calcium phosphate and its corresponding solubility product expression: Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq) Ksp = [Ca2+]3[PO43−]2 The molar solubility (S) of a slightly soluble salt is the number of moles of solute that dissolve in 1 L of solution and has the units M. For the example above, every 1 mol of solute that dissolves produces 3 mol of Ca2+ and 2 mol of PO43−, so the molar solubility is equivalent to S = [Ca2+]/3 = [PO43−]/2 If MgCrO4 were added to water to form 1 L of solution, the concentration of which species will be equal to the molar solubility? Check all that apply. CrO42− Mg2+ MgCrO4
David C.
What is the molar solubility of magnesium oxalate (MgC2O4) in water? The solubility product constant for MgC2O4 is 8.6x10^-5 at 25 degrees C. a. 4.3x10^-5 b. 9.3x10^-3 c. 1.3x10^-2 d. 1.7x10^-4 e. 8.35
Madhur L.
Question 1 : The solubility product, Ksp, of Cd3(PO4)2 is 2.5 x 10^-33. What is the solubility (in g/L) of Cd3(PO4)2 in pure water? Question 2 : The solubility product of Cu(OH)2 is 4.8 x 10^-20. Calculate the value of pCu2+, or -log[Cu2+], in an aqueous solution of NaOH which has a pH of 12.39 and is saturated in Cu(OH)2. Question 3 : The equilibrium constant for the formation of Cu(CN)4^2- is 2.0 x 10^30. Calculate the value of pCu2+, or -log[Cu2+], if we were to dissolve 2.87 g of CuCl2 in 1.000 L of a solution 0.836 M in NaCN. The addition of CuCl2 does not alter the volume (the final volume is still 1.000 L).
Sri K.
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD