Question

The pH of a normal raindrop is $\mathbf{5}\mathbf{.}\mathbf{60}$. Compute the concentrations of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$(aq), ${\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{‐}}$(aq), and ${\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{‐}}$(aq) in this raindrop if the total concentration of dissolved carbonates is $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{5}}\mathit{m}\mathit{o}\mathit{l}\mathit{}\mathit{/}\mathit{L}$.

Short answer

The concentration of ${\mathrm{H}}_2{\mathrm{CO}}_3$is $8.2\times {10}^{‐6}mol\mathit{}\mathit{/}\mathit{}L$, the concentration of ${\mathrm{HCO}}_3^{‐}$ is $1.4\times {10}^{‐6}mol\mathit{}\mathit{/}\mathit{}L$, and the concentration of ${\mathrm{CO}}_3^{2‐}\mathrm{is}2.7\times {10}^{‐11}mol\mathit{}\mathit{/}\mathit{}L$ .

Step-by-step solution

Calculate concentration of H3O+

Let us first calculate the concentration of ${\mathrm{H}}_3{{\mathrm{O}}^{+}}_{}$in the solution from the given pH.

$$\begin{gathered} \mathrm{pH}=5.6 \\ -\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=5.6 \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=2.5\times {10}^{‐4}mol\mathit{}\mathit{/}\mathit{}L \end{gathered}$$

Finding equilibrium expression for first equilibrium.

Let us write the equilibrium reactions that are taking place in the system.

$$\begin{gathered} \frac{\left[{\mathrm{HCO}}_3^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=K_1 \\ \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=\frac{{\mathrm{K}}_1}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]} \\ \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=0.17 \end{gathered}$$

Finding equilibrium expression for second equilibrium.

The equilibrium expression for the second equilibrium is as follows:

$$\begin{gathered} \frac{\left[{\mathrm{CO}}_3^{2-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}={\mathrm{K}}_2 \\ \frac{\left[{\mathrm{CO}}_3^{2-}\right]}{\left[{\mathrm{HCO}}_3^{2-}\right]}=\frac{{\mathrm{K}}_2}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]} \\ \frac{\left[{\mathrm{CO}}_3^{2-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}=1.9\times {10}^{-6} \end{gathered}$$

Given concentration of carbonates.

It is given that the total concentration of dissolved carbonates is ${10}^{-5}mol\mathit{}\mathit{/}\mathit{}L$.

$\left[{\mathrm{CO}}_3^{2-}\right]+\left[{\mathrm{HCO}}_3^{-}\right]+\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]={10}^{-5}$

Divide the entire equation by $\left[{\mathrm{HCO}}_3^{2-}\right]$.

$\frac{\left[{\mathrm{CO}}_3^{2-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}+1+\frac{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}=\frac{{10}^{-5}}{\left[{\mathrm{HCO}}_3^{-}\right]}$

Concentration of HCO3-

Substitute the ratios obtained previously in the expression above.

$\left(1.9\times {10}^{-5}\right)+1+\frac{1}{0.17}=\frac{{10}^{-5}}{\left[{\mathrm{HCO}}_3^{-}\right]}$

Evaluate the equation to obtain the value of $\left[{\mathrm{HCO}}_3^{-}\right]$.

$\left[{\mathrm{HCO}}_3^{-}\right]=1.4\times {10}^{-6}mol/L$

Concentration of H2CO3

Substitute the value of $\left[{\mathrm{HCO}}_3^{-}\right]$in first equilibrium equation to obtain the concentration of ${\mathrm{H}}_2{\mathrm{CO}}_3$.

$$\begin{gathered} \frac{1.4\times {10}^{-6}}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}=0.17 \\ \left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]=8.2\times {10}^{-6}mol/L \end{gathered}$$

Calculating the value of  HCO3-

Substitute the value of $\left[{\mathrm{HCO}}_3^{-}\right]$in second equilibrium equation to obtain the concentration of ${\mathrm{CO}}_3^{2-}$.

$$\begin{gathered} \frac{\left[{\mathrm{CO}}_3^{2-}\right]}{1.4\times {10}^{-6}}=1.9\times {10}^{-6} \\ \left[{\mathrm{CO}}_3^{2-}\right]=2.7\times {10}^{-11}mol/L \end{gathered}$$

Hence,

$$\begin{gathered} \left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]=8.2\times {10}^{-6}\mathit{}mol\mathit{}\mathit{/}\mathit{}L \\ \left[{\mathrm{HCO}}_3^{-}\right]=1.4\times {10}^{-6}mol\mathit{}\mathit{/}\mathit{}L \\ \left[{\mathrm{CO}}_3^{2-}\right]=2.7\times {10}^{-11}mol\mathit{}\mathit{/}\mathit{}L \end{gathered}$$

The concentration of ${\mathrm{H}}_2{\mathrm{CO}}_3\mathrm{is}8.2\times {10}^{-6}mol\mathit{}\mathit{/}\mathit{}L$, the concentration of ${\mathrm{HCO}}_3^{-}\mathrm{is}1.4\times {10}^{-6}mol\mathit{}\mathit{/}\mathit{}L$ , and the concentration of ${\mathrm{CO}}_3^{2-}\mathrm{is}2.7\times {10}^{-11}mol\mathit{}\mathit{/}\mathit{}L$ .