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Conversion of Solubility to Ksp

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Conversion of Solubility to Ksp

How is baking soda made?

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  K_{sp}

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:

\frac{4.2 \times 10^{-4} \ \cancel{\text{g}}}{\text{L}} \times \frac{1 \ \text{mol}}{99.41 \ \cancel{\text{g}}}=4.2 \times 10^{-6} \ \text{mol/L} \ (\text{M})

Solubility data can be used to calculate the  K_{sp} for a given compound. The following steps need to be taken.

  1. Convert form solubility to molar solubility.
  2. Use the dissociation equation to determine the concentration of each of the ions in mol/L.
  3. Apply the K_{sp} equation.

Sample Problem: Calculating  K_{sp} from Solubility

The solubility of lead(II) is found experimentally to be 0.533 g/L. Calculate the  K_{sp} 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

  • K_{sp} of PbF 2 = ?

The dissociation equation for PbF 2 and the corresponding  K_{sp} expression

\text{PbF}_2(s) \rightleftarrows \text{Pb}^{2+}(aq)+2\text{F}^-(aq) && K_{sp}=[\text{Pb}^{2+}][\text{F}^-]^2

The steps above will be followed to calculate the  K_{sp} for PbF 2 .

Step 2: Solve .

\text{molar solubility} \qquad \frac{0.533 \ \cancel{\text{g}}}{\text{L}} \times \frac{1 \ \text{mol}}{245.20 \ \cancel{\text{g}}}=2.17 \times 10^{-3} \ \text{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:

[\text{Pb}^{2+}]=2.17 \times 10^{-3} \ \text{M} \quad \text{and} \quad [\text{F}^-]=2 \times 2.17 \times 10^{-3}=4.35 \times 10^{-3} \ \text{M}

Substitute into the expression and solve for the K_{sp} .

K_{sp}=(2.17 \times 10^{-3})(4.35 \times 10^{-3})^2=4.11 \times 10^{-8}

Step 3: Think about your result .

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  K_{sp} using molar solubility are described.

Practice

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

http://www.chemteam.info/Equilibrium/Calc-Ksp-FromMolSolub.html

Review

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

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