Question 22.11: Consider a mixture of Cd(OH)2(s) and Cu(OH)2(s), for which t......

Let’s begin by considering the Cd(OH)_{2}(s). The dissolution equation and corresponding K_{sp} expression for Cd(OH)_{2}(s) are

and

The solubility of Cd(OH)_{2}(s) in water can be calculated from the K_{sp} expression:

s = [Cd^{2+}] =\frac{7.2 × 10^{–15}\ M^3}{[OH^–]^2}                (22.26)

The concentration of OH^−(aq) can be related to [H_{3}O^+] by using the ion product constant expression for water:

Substitution of this equation into Equation 22.26 yields

s = [Cd^{2+}] =\frac{7.2 × 10^{–15}\ M^3\ [H_{3}O^+]^2}{(1.0 × 10^{–14} M^2)^2}                          (22.27)

Now let’s consider the Cu(OH)_{2}(s), for which we have

and

We can find the solubility of Cu(OH)_{2}(s) in a manner analogous to that for Cd(OH)_{2}(s) by using the K_{sp} expression for Cu(OH)_{2}(s) and ion product concentration expression of water to obtain

s = [Cu^{2+}] =\frac{2.2 × 10^{–20}\ M^3}{[OH^–]^2} =\frac{2.2 × 10^{–20}\ M^3\ [H_{3}O^+]^2}{(1.0 × 10^{-14}\ M^{2})^{2}}              (22.28)

From Equations 22.27 and 22.28, we can calculate the solubility of Cd(OH)_{2}(s) and Cu(OH)_{2}(s) at various pH values, as shown in Table 22.4 and Table 22.5. These results are plotted in Figure 22.13. From Figure 22.13 we see that Cd(OH)_{2}(s) can be separated from Cu(OH)_{2}(s) by adjusting the pH of the solution to about 6.5 using a buffer solution. At pH ≈ 6.5, the Cd^{2+}(aq) dissolves, but the Cu(OH)_{2}(s) remains in solution.