When 75.0 mL of a 0.100 M lead(II) nitrate solution is mixed with 100.0 mL of a 0.190 M potass

step 1 We are asked to consider the reaction I've written here. Equus lead nitrate reacts with potassiu

When 75.0 mL of a 0.100 M lead(II) nitrate solution is...

When $75.0 \mathrm{~mL}$ of a $0.100 \mathrm{M}$ lead(II) nitrate solution is mixed with $100.0 \mathrm{~mL}$ of a $0.190 \mathrm{M}$ potassium iodide solution, a yelloworange precipitate of lead(II) iodide is formed. (a) What mass in grams of lead(II) iodide is formed, assuming the reaction goes to completion? (b) What is the molarity of each of the ions $\mathrm{Pb}^{2+}, \mathrm{K}^{+}, \mathrm{NO}_{3}^{-}$, and I in the resulting solution?

When $75.0 \mathrm{~mL}$ of a $0.100 \mathrm{M}$ lead(II) nitrate solution is mixed with $100.0 \mathrm{~mL}$ of a $0.190 \mathrm{M}$ potassium iodide solution, a yelloworange precipitate of lead(II) iodide is formed.
(a) What mass in grams of lead(II) iodide is formed, assuming the reaction goes to completion?
(b) What is the molarity of each of the ions $\mathrm{Pb}^{2+}, \mathrm{K}^{+}, \mathrm{NO}_{3}^{-}$, and I in the resulting solution?

Key Concepts

Molarity Calculations

Molarity calculations involve determining the concentration of a solute in a given volume of solution, typically expressed in moles per liter (M). This concept is used to compute the number of moles of reactants provided and the final concentration of various ions in a mixed solution after a reaction or dilution.

Stoichiometry

Stoichiometry involves using the balanced chemical equation to determine the mole ratios between reactants and products. This concept is crucial for calculating the amounts of substances that react or are produced in a chemical reaction, based on their molar relationships.

Limiting Reactant

The limiting reactant is the substance in a reaction that is completely consumed first, thereby determining the maximum amount of product that can form. Identifying the limiting reactant is essential for predicting the final yield of a product in any chemical reaction.

Precipitation Reactions

Precipitation reactions involve the formation of an insoluble product when two aqueous solutions are mixed. Understanding the solubility rules allows one to predict when a precipitate will form, as well as calculate the amount of the precipitate produced by the reaction.

Transcript

00:02 We are asked to consider the reaction i've written here.

00:05 Equus lead nitrate reacts with potassium iodide and the potassium, excuse me, the lead -2 iodide precipitates.

00:16 We're given the following of information.

00:19 We have 75 .0 milliliters of 0 .10 molar lead to nitrate.

00:34 We mix it with 100 milliliters of 0 .190 molar potassium iodide.

00:48 We are asked for part a, the mass of lead -2 iodide formed.

01:03 And for part b, we are asked the concentration of pb2 plus at the end, the k -plus, the nl3 minus, and the eye concentrations.

01:24 We need to know the malarity of these four species.

01:30 This is actually not as bad as i thought of it was going to be.

01:33 So this is limiting reactant problem.

01:37 And let's start by looking at our lead iodide, lead two iodide, lead to nitrate, i should say.

01:54 We're going to take our volume and liters times our malarity, and then i'm going to convert that to mass of our lead to iodide that will form.

02:26 It's a one -to -one ratio.

02:28 In the mold of mass my lead iodide was 461 .0 .01 grams per mole.

02:57 This equals 3 .45, i can't read my writing here, 3 .4576.

03:14 We'll worry about rounding later.

03:23 Then let's figure out how many moles or how much would be formed for the k -i.

03:34 And here we've got 100 milliliters, which is 0 .1 liters, 0 .190 moles per liter.

03:56 Here we'll have a 2 to 1 mole ratio, and our molar mass is still 4601 grams per mole, and this equals 4 .3796 grams of pb .i 2.

04:31 We can say that this is limiting and this will be the mass of lead iodide.

04:38 So the answer for a to three significant figures will be 3 .46 grams of lead -2 iodide will be formed.

04:55 And we can start looking at part b.

05:00 For a lead -2 iodide, or a lead -2, all of our lead -2, since that's limiting, will be converted to the lead -2 iodide, so our concentration will be zero.

05:17 Then let's do what's next.

05:22 I did n .o3 next...

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