A solution contains 9.54x10° M sodium chromate and 1.40x102 M ammonium sulfate. Solid lead nitrate is added slowly to this mixture. What is the concentration of chromate ion when sulfate ion begins to precipitate? [chromate] = 2.48x10^-7 x M Incorrect When lead nitrate is added to a solution of sodium chromate and ammonium sulfate two precipitates can form, lead chromate and lead sulfate. Notice that they have lead ion in common. Step 1: From the PBSO, equilibrium calculate the concentration of Pb²* needed to precipitate 1.40×102 M of SO,. PbSO4 (s) 2 Pb2+ (aq) + so, (aq) ? M 1.40×10-2 some Ksp = [Pb?] [SO,²] = 1.80×10 Ksp 1.80×10-8 [РЬ* - = 1.29×10“ M Pb2+ [SO,²] (1.40×10-2) Step 2: Use this concentration of Pb?* to calculate the concentration of CrO,² left in solution. PbCrO4 (s) 2 РЬ* (аq) + Cro, (aq) 1.29x106 M some Previous Next Step 2: Use this concentration of Pb²+ to calculate the concentration of Cro, 1left in solution. РЬСrО, (s) 2 РЬ* (aq) + Cro,- (aq) some 1.29×10-6 M ? M K = [Pb2] [CrO,] = 1.80×10-14 Ksp [Cro,?] = ( (РЬ) 1.80×10-14 = 1.40×10* M Cro,²- (1.29×106) Step 3: Compare the amount left of CrO,²" in solution to the initial amount: [Cro,?]o = 9.54×10-3 M > 1.40×10** M So PbCrO, precipitates until the [CrO,²] equals 1.40×10*8 M.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question:

A solution contains 5.90×10^-3 M ammonium carbonate and 1.11×10^-2 M ammonium fluoride.
Solid lead acetate is added slowly to this mixture.

What is the concentration of carbonate ion when fluoride ion begins to precipitate?
[carbonate] = M

A solution contains 9.54x10° M sodium chromate and 1.40x102 M ammonium sulfate.
Solid lead nitrate is added slowly to this mixture.
What is the concentration of chromate ion when sulfate ion begins to precipitate?
[chromate] = 2.48x10^-7 x M
Incorrect
When lead nitrate is added to a solution of sodium chromate and ammonium sulfate two precipitates can form, lead chromate and lead
sulfate. Notice that they have lead ion in common.
Step 1:
From the PBSO, equilibrium calculate the concentration of Pb²* needed to precipitate 1.40×102 M of SO,.
PbSO4 (s) 2 Pb2+ (aq) +
so, (aq)
? M
1.40×10-2
some
Ksp = [Pb?*] [SO,²] = 1.80×10*
Ksp
1.80×10-8
[РЬ* -
= 1.29×10“ M Pb2+
[SO,²]
(1.40×10-2)
Step 2:
Use this concentration of Pb?* to calculate the concentration of CrO,² left in solution.
PbCrO4 (s) 2
РЬ* (аq) + Cro, (aq)
1.29x106 M
some
Previous
Next

Step 2:
Use this concentration of Pb²+ to calculate the concentration of Cro, 1left in solution.
РЬСrО, (s) 2
РЬ* (aq) + Cro,- (aq)
some
1.29×10-6 M
? M
K = [Pb2*] [CrO,] = 1.80×10-14
Ksp
[Cro,?] = (
(РЬ*)
1.80×10-14
= 1.40×10* M Cro,²-
(1.29×106)
Step 3:
Compare the amount left of CrO,²" in solution to the initial amount:
[Cro,?]o = 9.54×10-3 M > 1.40×10** M
So PbCrO, precipitates until the [CrO,²] equals 1.40×10*8 M.

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