# Solubility and solubility product

• brake4country
In summary, the solubility product for Cu(OH)2 is represented by the expression Ksp=[Cu2+][OH-]2. This is due to the stoichiometry of the dissolution reaction, where for every one mole of Cu(OH)2 dissolved, 2 moles of OH- are produced. Therefore, the concentration of OH- is twice the solubility of Cu(OH)2. This is consistent with answer choice A, as the coefficients in the equilibrium expression are equal to the stoichiometric coefficients in the reaction.
brake4country

## Homework Statement

Which of the following expressions represents the solubility product for Cu(OH)2?
(A) Ksp=[Cu2+][OH-]2
(B) Ksp=[Cu2+]2[OH-]
(C) Ksp=[Cu2+]2[OH-]2
(D) Ksp=[Cu2+][OH-]

Ksp= [A][ B]

## The Attempt at a Solution

Okay, so I understand equilibrium expressions and in this case, solids and pure liquids are left out of the expression. My question is regarding the coefficients and exponents of solubility expressions. In chemical equilibrium expression is the concentration of the products are over the reactants (with exponents being the coefficients of each substance). Knowing that solubility rules are pretty much the same, I know that answer A is correct but my books reasoning for the following reaction does not make sense:

BaF2↔Ba2+aq+2F-aq
=Ksp=[Ba2+][F-]2

However, they write the equation as:

2.4 x 10-5 = (x)(2x)2 (and then they solve for x).

This expression is not consistent, say, with answer choice A in another problem. How did they get 2x in this instance for BaF2? Shouldn't the first problem above be Ksp=[Cu2+][2OH]2?[/B][/B][/B]

Last edited by a moderator:
Assuming x moles of BeF2 dissolved in 1 L of water, what is the concentration of Be2+? What is the concentration of F-? Can you plug them into Ksp?

We know that BeF2 dissociates to 1 mol of Be2+ and 2 mol of F-. but why must the coefficient of the F- be both 2x and squared? In equilibrium expressions and assuming that the reaction is elementary, the coefficients are the squares but not both. I hope I am making sense here.

You are squaring CONCENTRATION. Concentration is 2x, so its square is (2x)2.

Ok so in the original question with copper II hydroxide, answer A is wrong because we get 2OH for dissociation. Yes?

Nope, answer A is the correct one.

You are confused about what [OH-] is and what x is. They are not the same thing.

[OH-] is just the concentration of the OH-.

x is the solubility of the copper hydroxide. Concentration of OH- is not equal to the solubility, as I explained to you in my previous post. Concentration of OH- is twice the solubility of the copper hydroxide, as per each mole of copper hydroxide dissolved there are two moles of OH- produced. This is a simple result of the dissolution reaction stoichiometry:

Cu(OH)2 ↔ Cu2+ + 2OH-

Ok, thank you. I think my head is finally wrapping around this concept. What I was misinterpreting was the concentration of the hydroxide. For every one mole of Ca2+ there are actually 2 moles of OH-.

## 1. What is solubility?

Solubility refers to the ability of a substance, known as the solute, to dissolve in a solvent and form a homogeneous mixture. It is typically measured in terms of the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature and pressure.

## 2. How is solubility determined?

Solubility can be determined experimentally by adding a known amount of solute to a solvent and observing the amount that dissolves. It can also be predicted using mathematical equations such as the solubility product constant (Ksp) or by consulting solubility tables.

## 3. What factors affect solubility?

The solubility of a substance is affected by several factors including temperature, pressure, and polarity. In general, solubility increases with increasing temperature and decreasing pressure. Additionally, substances with similar polarities tend to be more soluble in each other.

## 4. What is solubility product (Ksp)?

Solubility product, also known as the solubility product constant, is a measure of the equilibrium concentration of ions in a saturated solution. It is equal to the product of the molar concentrations of ions raised to their respective stoichiometric coefficients in the dissociation equation of a slightly soluble compound.

## 5. How does solubility product relate to precipitation?

Solubility product can be used to predict whether a precipitation reaction will occur. If the product of the ion concentrations in a solution exceeds the solubility product, the solution is considered supersaturated and precipitation will occur. If the product is less than the solubility product, the solution is unsaturated and no precipitation will occur.

• Biology and Chemistry Homework Help
Replies
3
Views
2K
• Biology and Chemistry Homework Help
Replies
1
Views
2K
• Biology and Chemistry Homework Help
Replies
1
Views
2K
• Biology and Chemistry Homework Help
Replies
1
Views
2K
• Biology and Chemistry Homework Help
Replies
4
Views
8K
• Biology and Chemistry Homework Help
Replies
11
Views
6K
• Biology and Chemistry Homework Help
Replies
3
Views
3K
• Biology and Chemistry Homework Help
Replies
1
Views
5K
• Biology and Chemistry Homework Help
Replies
37
Views
5K
• Biology and Chemistry Homework Help
Replies
5
Views
2K