Question

Neglecting activity coefficients, calculate the pH of$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}{\mathbf{MHCIO}}_{\mathbf{4}}$What fraction of ${\mathbf{H}}^{\mathbf{+}}$is derived from dissociation of water?

Short answer

The pH value of $5.0\times {10}^{-8}{\mathrm{MHCIO}}_4=6.89$

The fraction of ${\mathrm{H}}^{+}$derived from dissociation of water = 0.61

Step-by-step solution

Define pH

pH stands for Potential hydrogen. It is the symbol for the logarithm of the reciprocal of hydrogen ion concentration in gram atoms per litre, used to express the acidity and alkalinity of a solution on a scale of $0\mathrm{to}14$

 Step 2: Determining the charge balance and mass balance 

Charge balance:$\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-1}\right]+\left[{{\mathrm{CIO}}_4}^{-}\right]\Rightarrow \left[{\mathrm{OH}}^{-}\right]=\left[{\mathrm{H}}^{+}\right]-5\times {10}^{-8}$

mass balance = charge balance

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}} \\ \left[{\mathrm{H}}^{+}\right]\left(\left[{\mathrm{H}}^{+}\right]-5\times {10}^{-8}\right)-1\times {10}^{14} \\ \left[{\mathrm{H}}^{+}\right]=1.28\times {10}^{-7} \end{gathered}$$

Calculating the pH value

Substituting the value of${\mathrm{H}}^{+}$ in the pH formula we get

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=6.89$

Calculating the fraction of total H+ from dissociation of water

$\left[{\mathrm{OH}}^{-}\right]={\mathrm{K}}_{\mathrm{w}}/\left[{\mathrm{H}}^{+}\right]=7.8\times {10}^{-8}\mathrm{M}$

Fraction of total $\left[{\mathrm{H}}^{+}\right]$from role="math" localid="1654777778226" ${\mathrm{H}}_2\mathrm{O}=\frac{7.8\times {10}^{-8}\mathrm{M}}{1.28\times {10}^{-7}\mathrm{M}}=0.61$