6.5 Precipitation and dissolution  (Page 3/17)

Determination of molar solubility from K _sp

The K _sp of copper(I) bromide, CuBr, is 6.3 $\times$ 10 ^–9 . Calculate the molar solubility of copper bromide.

Solution

The solubility product constant of copper(I) bromide is 6.3 $\times$ 10 ^–9 .

The reaction is:

First, write out the solubility product equilibrium constant expression:

Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the K _sp :

At equilibrium:

Therefore, the molar solubility of CuBr is 7.9 $\times$ 10 ^–5 M .

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The K _sp of AgI is 1.5 $\times$ 10 ^–16 . Calculate the molar solubility of silver iodide.

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Determination of molar solubility from K _sp , part ii

The K _sp of calcium hydroxide, Ca(OH) ₂ , is 1.3 $\times$ 10 ^–6 . Calculate the molar solubility of calcium hydroxide.

Solution

The solubility product constant of calcium hydroxide is 1.3 $\times$ 10 ^–6 .

The reaction is:

First, write out the solubility product equilibrium constant expression:

Create an ICE table, leaving the Ca(OH) ₂ column empty as it is a solid and does not contribute to the K _sp :

At equilibrium:

Therefore, the molar solubility of Ca(OH) ₂ is 1.3 $\times$ 10 ^–2 M .

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The K _sp of PbI ₂ is 1.4 $\times$ 10 ^–8 . Calculate the molar solubility of lead(II) iodide.

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Determination of K _sp From gram solubility

Many of the pigments used by artists in oil-based paints ( [link] ) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO ₄ , is 4.6 $\times$ 10 ^–6 g/L. Determine the solubility product equilibrium constant for PbCrO ₄ .

Solution

We are given the solubility of PbCrO ₄ in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb ²⁺ and ${\text{CrO}}_4{}^{\text{2−}},$ then K _sp :

1. Use the molar mass of PbCrO ₄ $\left(\frac{323.2\text{g}}{1\text{mol}}\right)$ to convert the solubility of PbCrO ₄ in grams per liter into moles per liter:

   $\begin{array}{}\\ \\ \left[{\text{PbCrO}}_4\right]=\frac{4.6\times {10}^{-6}{\text{g PbCrO}}_4}{1\text{L}}\times \frac{1{\text{mol PbCrO}}_4}{323.2{\text{g PbCrO}}_4}\\ =\frac{1.4\times {10}^{-8}{\text{mol PbCrO}}_4}{1\text{L}}\\ =1.4\times {10}^{-8}M\end{array}$

2. The chemical equation for the dissolution indicates that 1 mol of PbCrO ₄ gives 1 mol of Pb ²⁺ (aq) and 1 mol of ${\text{CrO}}_4{}^{\text{2−}}(aq)\text{:}$

   ${\text{PbCrO}}_4(s)\rightleftharpoons {\text{Pb}}^{\text{2+}}(aq)+{\text{CrO}}_4{}^{\text{2−}}(aq)$

   Thus, both [Pb ²⁺ ] and $[{\text{CrO}}_4{}^{\text{2−}}]$ are equal to the molar solubility of PbCrO ₄ :

   $[{\text{Pb}}^{\text{2+}}]=[{\text{CrO}}_4{}^{\text{2−}}]=\text{1.4}\times {10}^{-8}M$

3. Solve. K _sp = [Pb ²⁺ ] $[{\text{CrO}}_4{}^{\text{2−}}]$ = (1.4 $\times$ 10 ^–8 )(1.4 $\times$ 10 ^–8 ) = 2.0 $\times$ 10 ^–16

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The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.46 grams per liter at 20 °C. What is its solubility product?

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