Question

At $\mathbf{40}\mathbf{°}\mathbf{C}$and 1.00 atm pressure, a gaseous monoprotic acid has a density of $\mathbf{1}\mathbf{.}\mathbf{05} \mathbf{g}\mathbf{.}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}$. After 1.85 g of this gas is dissolved in water and diluted to 0 mL, the pH is measured to be 5.01. Determine the ${\mathbf{K}}_{\mathbf{3}}$ of this acid and use Table 15.2 to identify it.

Short answer

The ${\mathrm{K}}_{\mathrm{a}}$of acid is$6.27\times {10}^{-10}$.

Step-by-step solution

Volume of the acid

First, we calculate the volume of acid:

$$\begin{gathered} \rho =\frac{m}{V} \\ \mathrm{V}=\frac{\mathrm{m}}{\mathrm{\rho }} \\ \mathrm{V}=\frac{1.85\mathrm{g}}{1.05\mathrm{g}/\mathrm{L}} \\ \mathrm{V}=1.76\mathrm{L} \end{gathered}$$

The number of moles of the acid

Using the ideal gas law, we can calculate moles of acid:

$$\begin{gathered} \mathrm{pV}=\mathrm{nRT} \\ \mathrm{n}=\frac{\mathrm{pV}}{\mathrm{RT}} \\ \mathrm{n}=\frac{101325\mathrm{Pa}·1.76\times {10}^{-3}{\mathrm{m}}^3}{8.314\mathrm{J}/\mathrm{K}·\mathrm{mol}\times 313\mathrm{K}} \\ \mathrm{n}=0.06853\mathrm{mol} \end{gathered}$$

The concentration of the acid

The concentration of acid is:

$$\begin{gathered} \left[\mathrm{HA}\right]=\frac{\mathrm{n}}{\mathrm{V}} \\ \left[\mathrm{HA}\right]=\frac{0.06853\mathrm{mol}}{0.45\mathrm{L}} \\ \left[\mathrm{HA}\right]=0.15229\mathrm{M} \end{gathered}$$

Concentration of proton

The concentration of ${\mathrm{H}}^{+}$is:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}}={10}^{-5.01} \\ \left[{\mathrm{H}}^{+}\right]=9.77\times {10}^{-6}\mathrm{M} \end{gathered}$$

The concentration of $A^{-}$is equal to the concentration of ${\mathrm{H}}^{+}$:

$\left[{\mathrm{A}}^{-}\right]=\left[{\mathrm{H}}^{+}\right]$

The value of Ka

The ${\mathrm{K}}_{\mathrm{a}}$is:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{A}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{HA}\right]} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\left(9.77\times {10}^{-6}\right)}^2}{0.15229} \\ {\mathrm{K}}_{\mathrm{a}}=6.27\times {10}^{-10} \end{gathered}$$

According to Table 15.2, this is hydrocyanic acid (HCN).

Hence, the ${\mathrm{K}}_{\mathrm{a}}$of the acid is$6.27\times {10}^{-10}$.