Question

At \(25 ^\circ\text{C}\), \(K_a\) of formic acid is \(1.8 \times 10^{-4}\). What is the \(K_b\) of \( \text{HCOO}^- \)?

• A. \( 1.8 \times 10^{-10} \)
• B. \( 5.55 \times 10^{-4} \)
• C. \( 5.55 \times 10^{-11} \)
• D. \( 5.55 \times 10^{-12} \)

Hint:

For a conjugate acid-base pair (HA and A\(^-\)): \( K_a (\text{for HA}) \times K_b (\text{for A}^-) = K_w \). At \(25 ^\circ\text{C}\), the ionic product of water \( K_w = 1.0 \times 10^{-14} \). Given \(K_a\), find \(K_b\) using \( K_b = K_w / K_a \).

Answer 1

The Correct Option is C

Formic acid (HCOOH) is a weak acid.
Its conjugate base is formate ion (HCOO\(^-\)).
For a conjugate acid-base pair, the relationship between the acid dissociation constant (\(K_a\)) of the acid and the base dissociation constant (\(K_b\)) of its conjugate base is given by: \[ K_a \times K_b = K_w \] where \(K_w\) is the ionic product of water.
At \(25 ^\circ\text{C}\), \( K_w = 1.
0 \times 10^{-14} \).
Given \( K_a \) for formic acid = \( 1.
8 \times 10^{-4} \).
We need to find \( K_b \) for HCOO\(^-\).
\[ K_b = \frac{K_w}{K_a} = \frac{1.
0 \times 10^{-14}}{1.
8 \times 10^{-4}} \] \[ K_b = \frac{1.
0}{1.
8} \times \frac{10^{-14}}{10^{-4}} = \frac{1}{1.
8} \times 10^{-14 - (-4)} = \frac{1}{1.
8} \times 10^{-10} \] Calculate \( \frac{1}{1.
8} \): \( \frac{1}{1.
8} = \frac{10}{18} = \frac{5}{9} \).
\( 5 \div 9 = 0.
5555.
.
.
\) So, \( \frac{1}{1.
8} \approx 0.
555 \).
\[ K_b = 0.
555 \times 10^{-10} = 5.
55 \times 10^{-1} \times 10^{-10} = 5.
55 \times 10^{-11} \] This matches option (3).