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.
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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Question
A solution contains 1.32×10-2 M silver nitrate and 6.56×10-3 M lead acetate.
Solid potassium iodide is added slowly to this mixture.
What is the concentration of silver ion when lead ion begins to precipitate?
[Ag+] = M
Attached photo of how to solve.
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7085239-dcca-4c24-8a3e-855ec5c43846%2F487b8e07-cf08-49d0-856c-b99cc0b59a2c%2F2xmzoto.png&w=3840&q=75)
Transcribed Image Text: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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7085239-dcca-4c24-8a3e-855ec5c43846%2F487b8e07-cf08-49d0-856c-b99cc0b59a2c%2Fap14e8h.png&w=3840&q=75)
Transcribed Image Text: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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