Question

Using activities, calculate the pH and concentration of ${\mathbf{H}}^{\mathbf{+}}$in 0.050MLiBr at$\mathbf{25}\mathbf{°}\mathbf{C}$.

Short answer

The pH of the given solution is 6.99

Step-by-step solution

Define the formula for PH calculation.

The pH of a solution can be calculated as,

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}{\mathit{A}}_{{\mathbf{H}}^{\mathbf{+}}} \\ \mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\left[H^{+}\right]{\mathbf{\gamma }}_{{\mathbf{H}}^{\mathbf{+}}} \end{gathered}$$

where,

${\mathit{A}}_{{\mathbf{H}}^{+}}$ - activity of hydrogne ion

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

Calculate the PH form the activity.

Given

0.050MLiBr at $25°\mathrm{C}$

At an ionic strength role="math" localid="1654765585845" $\left(\mu \right)$of and${\mathrm{\gamma }}_{\mathrm{OH}}=0.81$.

role="math" localid="1654765776290" $$\begin{gathered} \left[{\mathrm{H}}^{+}\right]{\gamma }_{{\mathrm{H}}^{+}}\left[{\mathrm{OH}}^{-}\right]{\gamma }_{\mathrm{OH}} \\ =\left(\mathrm{x}\right)\left(0.86\right)\left(\mathrm{x}\right)\left(0.81\right) \\ =1.0\times {10}^{-14} \\ \mathrm{x}=1.2\times {10}^{-7}\mathrm{M}. \end{gathered}$$

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

Calculate the pH by using the concentration of ${\mathrm{H}}^{+}$ion

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

Thus, theof the given solution is 6.99 .