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

In summary, the conversation is about a student seeking help with two chemistry questions involving solubility and Ksp. The student shares their attempts at solving the questions and asks for further insight. The expert summarizes the conversation and suggests reaching out to the professor for clarification on the correct answers.

## 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!

## 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.

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.

## What is molar solubility?

Molar solubility refers to the maximum amount of a solute that can be dissolved in a given amount of solvent at a specific temperature and pressure.

## How is molar solubility calculated?

Molar solubility is calculated by dividing the number of moles of solute that can be dissolved by the volume of solvent in liters.

## What factors influence molar solubility?

The factors that influence molar solubility include temperature, pressure, and the nature of the solute and solvent.

## Why is molar solubility important in chemistry?

Molar solubility is important in chemistry because it helps determine the solubility of a substance and its ability to dissolve in a given solvent. This information is useful in various industrial and research applications.

## How does molar solubility differ from solubility?

Molar solubility differs from solubility in that it takes into account the amount of solute and solvent in a specific unit of measurement, while solubility is a more general term that refers to the ability of a substance to dissolve in a solvent.

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