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# Conversion of Solubility to Ksp

## Demonstrates how solubility constants can be derived from experimentally determined solubility.

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Conversion of Solubility to Ksp

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Baking soda (sodium bicarbonate) is prepared by bubbling carbon dioxide gas through a solution of ammonia and sodium chloride. Ammonium carbonate is first formed which then reacts with the NaCl to form sodium bicarbonate and ammonium chloride. The sodium bicarbonate is less soluble than the other materials, so it will precipitate out of solution.

### Conversion of Solubility to Ksp\begin{align*} K_{sp}\end{align*}

Solubility is normally expressed in g/L of saturated solution. However, solubility can also be expressed as the moles per liter. Molar solubility is the number of moles of solute in one liter of saturated solution. In other words, the molar solubility of a given compound represents the highest molarity solution that is possible for that compound. The molar mass of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of Zn(OH) 2 is 4.2 × 10 -4  g/L, the molar solubility can be calculated as shown below:

4.2×104 gL×1 mol99.41 g=4.2×106 mol/L (M)

Solubility data can be used to calculate the  Ksp\begin{align*}K_{sp}\end{align*} for a given compound. The following steps need to be taken.

1. Convert from solubility to molar solubility.
2. Use the dissociation equation to determine the concentration of each of the ions in mol/L.
3. Apply the Ksp\begin{align*}K_{sp}\end{align*} equation.

#### Sample Problem: Calculating  Ksp\begin{align*}K_{sp} \end{align*} from Solubility

The solubility of lead(II) fluoride is found experimentally to be 0.533 g/L. Calculate the  Ksp\begin{align*}K_{sp}\end{align*} for lead(II) fluoride.

Step 1: List the known quantities and plan the problem .

Known

• solubility of PbF 2 = 0.533 g/L
• molar mass = 245.20 g/mol

Unknown

• Ksp\begin{align*}K_{sp}\end{align*} of PbF 2 = ?

The dissociation equation for PbF 2 and the corresponding  Ksp\begin{align*}K_{sp}\end{align*} expression

PbF2(s)Pb2+(aq)+2F(aq)Ksp=[Pb2+][F]2

The steps above will be followed to calculate the  Ksp\begin{align*}K_{sp}\end{align*} for PbF 2 .

Step 2: Solve .

molar solubility0.533 gL×1 mol245.20 g=2.17×103 M

The dissociation equation shows that for every mole of PbF 2 that dissociates, 1 mol of Pb 2+ and 2 mol of F are produced. Therefore, at equilibrium the concentrations of the ions are:

[Pb2+]=2.17×103 Mand[F]=2×2.17×103=4.35×103 M

Substitute into the expression and solve for the Ksp\begin{align*}K_{sp}\end{align*} .

Ksp=(2.17×103)(4.35×103)2=4.11×108

The solubility product constant is significantly less than 1 for a nearly insoluble compound such as PbF 2 .

#### Summary

• Molar solubility calculations are described.
• Calculations of  Ksp\begin{align*}K_{sp}\end{align*} using molar solubility are described.

#### Practice

Read the material at the link below and do the problems at the end:

#### Review

Questions

1. What are the solution requirements for determining molar solubility?
2. Why do we need to convert mass to molarity to determine Ksp\begin{align*}K_{sp}\end{align*} ?
3. What  Ksp\begin{align*}K_{sp}\end{align*} values would you expect for very insoluble compounds?