Question

Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).

Answer 1

Answer: The pH of 0.10 M $Cu(NO_{3})_{2}(aq)$ is 4.49.

Explanation:

Given: Initial concentration of $Cu(H_{2}O)^{2+}_{6}$ = 0.10 M

$K_{a} = 1.0 \times 10^{-8}$

Let us assume that amount of $Cu(H_{2}O)^{2+}_{6}$ dissociates is x. So, ICE table for dissociation of  $Cu(H_{2}O)^{2+}_{6}$  is as follows.

                               $Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}$

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of $K_{a}$ is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for $K_{a}$ value is as follows.

$K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}$

Hence, $[H_{3}O^{+}] = 3.2 \times 10^{-5}$

Formula to calculate pH is as follows.

$pH = -log [H^{+}]$

Substitute the values into above formula as follows.

$pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49$

Thus, we can conclude that the pH of 0.10 M $Cu(NO_{3})_{2}(aq)$ is 4.49.