Correct Problem 7.121 When 78.0 mL of a 0.105 M lead (II) nitrate solution is mi

Question

Correct Problem 7.121 When 78.0 mL of a 0.105 M lead (II) nitrate solution is mixed with 120.0 mL of a 0.185 M potassium iodide solution, a yellow-orange precipitate of lead(II) iodide is formed Part C What is the molarity of K in the resulting solution? molarity = Submit My Answers Give Up Part D What is the molarity of NO3 in the resulting solution? No, molarity= Submit My Answers Give Up Part E What is the molarity of I in the resulting solution? molarity Submit My Answers Give Up Provide Feedback ContinuR

Explanation / Answer

The reaction will be---

Pb(NO3)2(aq) + 2KI(aq) ----> PbI2(s) + 2KNO3(aq)

Here in the above equation we can see that 1 mol Pb(NO3)2 reacts with 2 mol KI to produce 1 mol PbI2 and 2 mol KNO3 .

Now,
Mol Pb(NO3)2 in 78.0mL of 0.105 M solution = 0.0780 L x 0.105 M = 0.00819 mol Pb(NO3)2

This will react with = 0.00819 mol Pb(NO3)2 x (2 mol KI / 1 mol Pb(NO3)2 )=   0.01638 mol KI

Mol KI in 120 mL of 0.185 M solution = 0.120 L x 0.185 = 0.0222 mol KI

The final volume of solution = (78.0+120.0 ) mL = 198.0 mL = 0.198 L

The final solution will contain-----
Part C) K+ = 0.0222 mol K+ in 0.198 L solution:

Molarity of K+ = 0.0222 mol /0.198 L = 0.112 M

Part D) NO3- = 0.00819 x 2 = 0.01638 mol in 0.198 L solution

Molarity of NO3- = 0.01638 mol / 0.198 L = 0.0827 M

Part E) I- = (0.0222  - 0.01638) mol = 0.00582 mol I- in 0.198 L solution

Molarity of I- = 0.00582 mol /0.198 L = 0.0293 M

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