Questions: Calculate the pH for the following weak acid. A solution of HCOOH has 0.20 M HCOOH at equilibrium. The Ka for HCOOH is 1.8 x 10^-4. What is the pH of this solution at equilibrium? Express the pH numerically.

Solution Steps

The ionization of formic acid (\(\text{HCOOH}\)) in water can be represented as: \[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \]

The expression for the acid dissociation constant (\( K_a \)) is: \[ K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \] Given \( K_a = 1.8 \times 10^{-4} \).

Initially, the concentration of \(\text{HCOOH}\) is 0.20 M, and the concentrations of \(\text{H}^+\) and \(\text{HCOO}^-\) are 0 M.

Let \( x \) be the change in concentration at equilibrium:

• \([\text{HCOOH}] = 0.20 - x\)
• \([\text{H}^+] = x\)
• \([\text{HCOO}^-] = x\)

Substitute the equilibrium concentrations into the \( K_a \) expression: \[ 1.8 \times 10^{-4} = \frac{x \cdot x}{0.20 - x} \]

Assume \( x \) is small compared to 0.20, so \( 0.20 - x \approx 0.20 \): \[ 1.8 \times 10^{-4} = \frac{x^2}{0.20} \]

Solve for \( x \) (concentration of \(\text{H}^+\)): \[ x^2 = 1.8 \times 10^{-4} \times 0.20 \] \[ x^2 = 3.6 \times 10^{-5} \] \[ x = \sqrt{3.6 \times 10^{-5}} \] \[ x \approx 6.0 \times 10^{-3} \]

The pH is calculated using the formula: \[ \text{pH} = -\log[\text{H}^+] \] \[ \text{pH} = -\log(6.0 \times 10^{-3}) \] \[ \text{pH} \approx 2.22 \]

Final Answer