Question

Consider a solution formed by mixing 200.0 mL of0.250 M, 135.0 mL of 1.000 M , and 100.0 mL of 0.100 M .
a. Calculate the pH of this solution.
b. Calculate the concentration of$\mathbf{HCN}$ in this solution.

Short answer

1. The pH of this solution is$2.21$.
2. The concentration of $\mathrm{HCN}$in this solution $2.3\times {10}^{-2}\mathrm{M}$.

Step-by-step solution

Introduction to the concept

The following equation can be used to determine the pH of a solution.

$\mathbf{pH}\mathbf{=}\mathbf{–}\mathbf{log}\left[H^{+}\right]$

Determination of H3PO4 formed.

Number of moles of ${\mathrm{Na}}_3{\mathrm{PO}}_4=\frac{0.250\text{mol}}{1000\text{mL}}\times 200.0\text{mL}$

$=0.05\text{mol}$

Number of moles of $\mathrm{HCl}=\frac{1\text{mol}}{1000\text{mL}}\times 130.0\text{mL}$

$=0.135\text{mol}$

Number of moles of $NaCN=\frac{0.1\text{mol}}{1000\text{mL}}\times 100.0\text{mL}$

$=0.01\text{mol}$

$2{\mathrm{H}}^{+}+{\mathrm{PO}}_4^{3-}\to {\text{H}}_2{\mathrm{PO}}_4^{-}$

Number of moles of$\mathrm{HCl}$ reacts with${\mathrm{PO}}_4^{3-}$ to convert into${\text{H}}_2{\mathrm{PO}}_4^{-}=0.100\text{mol}$

${\mathrm{H}}^{+}+{\mathrm{CN}}^{-}\to \mathrm{HCN}$

The number of moles$\mathrm{HCl}$ reacts with ${\mathrm{CN}}^{-}=0.0010\text{mol}$.

The number of moles left$\mathrm{HCl}$ reacts $=0.025\text{mol}$.

${\mathrm{H}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\to {\text{H}}_3{\mathrm{PO}}_4$

Number of moles formed${\mathrm{H}}_3{\mathrm{PO}}_4$$=0.025\text{mol}$ .

Determination of pH formed

Finally, 0.01 mol of $\mathrm{HCN}$, 0.025 mol of ${\mathrm{H}}_2{\mathrm{PO}}_4$, and 0.025 mol of${\mathrm{H}}_3{\mathrm{PO}}_4$ remain in the solution.

Compared to${\mathrm{H}}_2{\mathrm{PO}}_4^{-}$ and ${\mathrm{H}}_3{\mathrm{PO}}_4^{-}$,$\mathrm{HCN}$ is a significantly weaker acid. As a result, the contribution of${\mathrm{H}}^{+}$ from$\mathrm{HCN}$ can be ignored.

The reaction's equilibrium constant will be:

${\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2\times \left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]}{\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]}$

$7.5\times {10}^{-3}={\mathrm{x}}^2+4.31\times {10}^{-14}$

$\mathrm{x}=0.0061\text{M}$

$\mathrm{pH}=2.21$

Therefore, the pH is $2.21$.

Determination of concentration of HCN. 

${\mathrm{K}}_{\mathrm{a}}\text{ofHCN=6.2}\times {\text{10}}^{-10}$

So,${\mathrm{K}}_{\mathrm{a}}$ is very small, because of this dissociation cannot be happened.

Thus, the concentration of$\mathrm{HCN}=\frac{0.010\text{mol}}{435\text{mL}}\times 1000\text{mL}$

$\text{=\hspace{0.33em}2.3}\times {\text{10}}^{-2}\text{M}$

Therefore, the concentration of$\mathbf{HCN}$ is $2.3\times {10}^{-2}M$.