(Without activities), calculate the pH of

(a) $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{MHBr}$

(b) role="math" localid="1654840353627" $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$$\mathbf{(}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$dissociates completely to $\mathbf{2}{\mathbf{H}}^{\mathbf{+}}$plus ${\mathbf{SO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$at this low concentration).

The number of all species present in a solution containing a specific atom should be equal to the quantity of that atom delivered to the solution, which is known as mass balance.

Charge balance is defined as the sum of all positive ions present in a solution equalling the sum of all negative ions present in the solution.

(a)

Given Charge balance of$\mathrm{HBr}=\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+\left[{\mathrm{Br}}^{-}\right]$

Mass balance of$\mathrm{HBr}=\left[{\mathrm{Br}}^{-}\right]=1.0\times {10}^{-8}\mathrm{M}$

Equilibrium$=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}$

Take

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\mathrm{x} \\ \left[{\mathrm{Br}}^{-}\right]=1.0\times {10}^{-8}\mathrm{M} \end{gathered}$$

From the above charge balance, $\left[{\mathrm{OH}}^{-}\right]=\mathrm{x}-1.0\times {10}^{-8}\mathrm{M}$

$\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}$

Put in$\left[{\mathrm{OH}}^{-}\right]=\mathrm{x}-1.0\times {10}^{-8}\mathrm{M}\mathrm{in}{\mathrm{K}}_{\mathrm{w}}$

(x)$$\begin{gathered} \left(\mathrm{x}-1.0\times {10}^{-8}\right)=1.0\times {10}^{-14} \\ \mathrm{x}=1.05\times {10}^{-7}\mathrm{M} \end{gathered}$$

Now pH will be.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left(1.0_5\times {10}^{-7}\mathrm{M}\right) \\ \mathrm{pH}=6.98 \end{gathered}$$

The pH of$1.0\times {10}^{-8}\mathrm{MHBr}=6.98$

(b)

Given Charge balance of${\mathrm{H}}_2{\mathrm{SO}}_4=\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]+2\left[{{\mathrm{SO}}_4}^{2-}\right]$

Mass balance of${\mathrm{H}}_2{\mathrm{SO}}_4=\left[{\mathrm{SO}}_4^{2-}\right]=1.0\times {10}^{-8}\mathrm{M}$

Equilibrium$=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}$

Take,

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\mathrm{x} \\ \left[{{\mathrm{SO}}_4}^{2-}\right]=1.0\times {10}^{-8}\mathrm{M} \end{gathered}$$

On charge balance$\left[{\mathrm{OH}}^{-}\right]=\mathrm{x}-1.0\times {10}^{-8}\mathrm{M}$

$\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}$

Substitute $\left[{\mathrm{OH}}^{-}\right]=\mathrm{x}-2.0\times {10}^{-8}\mathrm{M}\mathrm{in}{\mathrm{K}}_{\mathrm{w}}$

(x) role="math" localid="1654841691499" $\left(x-2.0\times {10}^{-8}\right)=1.0\times {10}^{-14}$

$\mathrm{x}=1.10\times {10}^{-7}\mathrm{M}$

Now the will be

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\mathrm{pH} \\ =-\log \left(1.10\times {10}^{-7}\mathrm{M}\right) \\ \mathrm{pH}=6.96 \end{gathered}$$

Thus, The pH of$1.0\times {10}^{-8}{\mathrm{MH}}_2{\mathrm{SO}}_4=6.96$