What is the Correct Molar Solubility for Barium Chromate and Silver Phosphate?

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The discussion revolves around calculating the solubility product constant (Ksp) for barium chromate (BaCrO4) and the molar solubility of silver phosphate (Ag3PO4). The calculated Ksp for BaCrO4 is approximately 3.6×10-10 based on its solubility of 1.9×10-5 mol/L. For Ag3PO4, using its Ksp of 1.8×10-18, the molar solubility was found to be around 1.6×10-5 mol/L. Participants express confusion over the calculations, particularly regarding algebraic steps. The original poster plans to contact their professor for clarification on the correct answers.
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Homework Statement


I have worked these two problems to the best of my knowledge and I keep getting the wrong answer (I'm not sure what the right answer is). Here are the questions:

1. At a certain temperature the solubility of barium chromate (BaCrO4) is 1.9×10-5 mol/L. What is the Ksp of BaCrO4 at this temperature?
2. What is the molar solubility of silver phosphate (Ag3PO4) in water? Ksp = 1.8×10-18.

Homework Equations



The relevant equations are Ksp=([products]/[reactants])

The Attempt at a Solution


For question 1: BaCrO4⇔Ba2+ + CrO42-
Ksp=[Ba2+][CrO42-]/1 (since it is a solid)
Ksp=[x][x]=x2
Ksp=[1.9x10-5]2=3.61x10-10 = 3.6x10-10

For question 2: Ag3PO4⇔3Ag+ + PO43-
1.8x10-18 = ([Ag+]3[PO43-])/1
1.8x10-18 = [3x]3[x]
1.8x10-18 = [27x4] then divide both sides by 27
6.6666667x10-20 = x4 then do the fourth root on both sides
x = 1.6068568x10-5 = 1.6x10-5

I would very much appreciate any insight into this. Thank you!
 
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mainzelmadchen said:

Homework Statement


I have worked these two problems to the best of my knowledge and I keep getting the wrong answer (I'm not sure what the right answer is). Here are the questions:

1. At a certain temperature the solubility of barium chromate (BaCrO4) is 1.9×10-5 mol/L. What is the Ksp of BaCrO4 at this temperature?
2. What is the molar solubility of silver phosphate (Ag3PO4) in water? Ksp = 1.8×10-18.


Homework Equations



The relevant equations are Ksp=([products]/[reactants])

The Attempt at a Solution


For question 1: BaCrO4⇔Ba2+ + CrO42-
Ksp=[Ba2+][CrO42-]/1 (since it is a solid)
Ksp=[x][x]=x2
Ksp=[1.9x10-5]2=3.61x10-10 = 3.6x10-10

For question 2: Ag3PO4⇔3Ag+ + PO43-
1.8x10-18 = ([Ag+]3[PO43-])/1
1.8x10-18 = [3x]3[x]
1.8x10-18 = [27x4] then divide both sides by 27
6.6666667x10-20 = x4 then do the fourth root on both sides
x = 1.6068568x10-5 = 1.6x10-5

I would very much appreciate any insight into this. Thank you!

Your approach looks fine to me unless you have done some calculation mistake.

What are the correct answers?
 
It is an online homework, like a quiz, that we are allowed multiple attempts on. It doesn't give me the correct answer. I'll email the prof. Thanks!
 
where is this coming from? [27x4]
 
Which part of (3x)^3x = 27x^4 is unclear to you? This is a pretty basic algebra.
 
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