Q19.21P - Chemistry: Molecular Nature Of Matter And Change

Question

A buffer consists of 0.22 M ${\mathbf{\text{KHCO}}}_{\mathbf{\text{3}}}$ and 0.37 M ${\mathbf{\text{K}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{3}}}$. Carbonic acid is a diprotic acid with ${\mathbf{\text{K}}}_{\mathbf{\text{a1}}}{\mathbf{\text{ = 4.5×10}}}^{\mathbf{-}\mathbf{7}}$ and ${\mathbf{\text{K}}}_{\mathbf{\text{a2}}}{\mathbf{\text{ = 4.7×10}}}^{\mathbf{-}\mathbf{11}}$. (a) Which  value is more important to this buffer? (b) What is the buffer pH?

Step-by-Step Solution

Answer

(a) The value important to this buffer is ${\text{K}}_{\text{a2}}$.

(b) The buffer pH is 10.57.

Step 1: Ionic Equilibrium

An ionic equilibrium refers to an equilibrium that exists in the solution of weak electrolytes between unionized molecules and ions.

Step 2: Subpart (a)

First, write the carbonic acid dissociation process.

$$\begin{gathered} {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{HCO}}_{\text{3}}^{\text{ - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }} \\ {\text{HCO}}_{\text{3}}^{\text{ - }}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CO}}_{\text{3}}^{\text{2 - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }} \end{gathered}$$

Write the potassium salt's dissolution.

$$\begin{gathered} {\text{KHCO}}_{\text{3}}\rightleftharpoons {\text{K}}^{\text{ + }}{\text{ + HCO}}_{\text{3}}^{\text{ - }} \\ {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\rightleftharpoons {\text{2K}}^{\text{ + }}{\text{ + CO}}_{\text{3}}^{\text{2 - }} \end{gathered}$$

${\text{HCO}}_{\text{3}}^{\text{ - }}$and  ${\text{CO}}_{\text{3}}^{\text{2 - }}$were both generated by salts. Both sides of the carbonic acid's second dissociation contain these chemical entities. Therefore, for this buffer solution, we'll utilize ${\text{K}}_{\text{a2}}$.

Step 3: Subpart (b)

The given bicarbonate and carbonate concentrations are:

$$\begin{gathered} {\text{[HCO}}_{\text{3}}^{\text{ - }}\text{]=0.22M} \\ {\text{[CO}}_{\text{3}}^{\text{2 - }}\text{]=0.37M} \end{gathered}$$

To find the pH, we may utilize the Henderson-Hasselbalch equation. First, use ${\text{K}}_{\text{a2}}$ to solve for the ${\text{pK}}_{\text{a}}$.

$$\begin{gathered} {\text{ pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(4.7×10}}^{-11}\text{)} \\ =10.34 \\ {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\text{[base]}}{\text{[ acid ]}} \\ {\text{ = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{CO}}_{\text{3}}^{\text{2 - }}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{ - }}\right]} \\ \text{ = 10.34 + log}\frac{\text{0.37M}}{\text{0.22M}} \\ \text{ = 10.57.} \end{gathered}$$

Therefore, the value of pH is 10.57.