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

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Discussion Overview

The discussion revolves around the calculation of molar solubility and solubility product constant (Ksp) for barium chromate (BaCrO4) and silver phosphate (Ag3PO4). Participants are attempting to solve homework problems related to these compounds, focusing on the application of Ksp equations and algebraic manipulation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents their calculations for the Ksp of BaCrO4 based on its solubility, concluding with a value of 3.6x10-10.
  • The same participant calculates the molar solubility of Ag3PO4, arriving at a value of 1.6x10-5, based on the provided Ksp of 1.8x10-18.
  • Another participant expresses uncertainty about the correctness of the answers and suggests that there might be a calculation mistake.
  • A later reply questions the algebraic step involving [27x4], seeking clarification on its derivation.
  • Another participant responds by asserting that the algebraic manipulation is basic and should be clear.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the calculated answers, as one participant expresses doubt about their accuracy while another believes the calculations are fine. The discussion remains unresolved regarding the correct answers.

Contextual Notes

Participants have not provided explicit assumptions or constraints that might affect the calculations, and the discussion does not clarify whether any specific conditions apply to the solubility values or Ksp equations used.

mainzelmadchen
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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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