Question

Question: A chemist dissolves 0.135 mole of CO₂(g) in 2.50 L of 0.105 MNa₂CO₃. Calculate the pH of the resulting solution.

Short answer

Answer:The pH of the resulting solution is 10.0.

Step-by-step solution

Determination of concentration of H2CO3

When CO₂ is dissolves in the solution, it is converted into H₂CO₃.

Concentration of H₂CO₃ is,

$\frac{0.135mol}{2.50L}=5.40\times {10}^{-2}M$

  Step 2: Determination of equilibrium

The chemical reaction for the process is,

$$\begin{gathered} H_{2}C{O}_3\left(aq\right)+C{O}_3^{2-}\left(aq\right)\to 2HC{O}_3^{-}\left(aq\right) \\ {K}_{a1}=\frac{\left[H^{+}\right]\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]} \\ {K}_{a2}=\frac{\left[H^{+}\right]\left[C{O}_3^{-2}\right]}{\left[HC{O}_3^{-}\right]} \\ \frac{K_{a1}}{K_{a2}}=\frac{\frac{\left[H^{+}\right]\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}}{\frac{\left[H^{+}\right]\left[C{O}_3^{-2}\right]}{\left[HC{O}_3^{-}\right]}} \\ \frac{K_{a1}}{K_{a2}}=\frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]\left[C{O}_3^{-2}\right]} \\ \frac{4.3\times {10}^{-7}}{4.8\times {10}^{-11}}=9.0\times {10}^3 \end{gathered}$$

From this, calculate the concentration of HCO₃^- and CO₃^2-,

$$\begin{gathered} 2\left(5.4\times {10}^{-2}\right)=0.108 M HC{O}_3^{-} \\ 0.105-5.4\times {10}^{-2}=0.051 M C{O}_3^{2-} \end{gathered}$$

Using the value of K_a1, calculate [H⁺],

  Step 3: Determination of pH

The pH of the solution is,

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ =-\log \left(1.0\times {10}^{-10}\right) \\ =10.0 \end{gathered}$$

Thus, the pH of the solution is 10.0.