Question

Find $\left[H^{+}\right]$ and the pH of$\mathbf{0}\mathbf{.}\mathbf{05}{\mathbf{MI}}^{\mathbf{2}}{\mathbf{NO}}_{\mathbf{3}}$.

Short answer

The concentration of ${\mathrm{H}}^{+}$ ion is $1.2\times {10}^{-7}\mathrm{M}$.

The pH is 6.99.

Step-by-step solution

Concept used.

pH of a solution:

The pH of a solution can be calculated by,

$\mathbf{pH}\mathbf{=-}{\mathbf{logA}}_{{\mathbf{H}}^{\mathbf{+}}}$

${\mathit{\gamma }}_{{\mathbf{H}}^{\mathbf{+}}}$-activity coefficient of hydrogen ion

With the help of above equation, the negative logarithm of activity of${\mathbf{H}}^{\mathbf{+}}$ ion can be calculated.

Step 2: Calculate the value of [H+] and the pH of .

At an ionic strength$\left(\mu \right)$of$0.050M,{\gamma }_{H^{+}}=0.86$and${\gamma }_{OH}=0.81$from the table 8 - 1.

role="math" localid="1663412820591" $$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]{\mathrm{\gamma }}_{{\mathrm{H}}^{+}}\left[{\mathrm{OH}}^{-}\right]{\mathrm{\gamma }}_{{\mathrm{OH}}^{-}} \\ =\left(\mathrm{x}\right)\left(0.86\right)\left(\mathrm{x}\right)\left(0.81\right) \\ =1.0\times {10}^{-14} \\ \mathrm{x}=\left[{\mathrm{H}}^{+}\right] \\ =1.2\times {10}^{-7}\mathrm{M} \end{gathered}$$

The concentration of $\left[{\mathrm{H}}^{+}\right]$ion is $1.2\times {10}^{-7}\mathrm{M}$.

By using the concentration and activity coefficient of ion calculate the pH

$$\begin{gathered} pH=-\log \left[\left(1.2\times {10}^{-7}\left)\right(0.86\right)\right] \\ =6.99 \end{gathered}$$

The pH is 6.99.