Question

Calculate the pH of the solution that results from the following mixtures.

50.0 mL of 0.17 molL⁻¹ HCOOH (Ka=1.8×10⁻⁴) with 80.0 mL of 0.12 molL⁻¹ HCOONa. Express your answer using two decimal places.

Answer 1

The pH of the solution resulting from the mixture of HCOOH and HCOONa is 3.79.

Step-by-step explanation:

To calculate the pH of the solution resulting from the mixture, we need to consider the dissociation of the weak acid, HCOOH, and the reaction with the strong base, HCOONa. First, we calculate the amount of HCOOH and HCOONa in moles:

HCOOH: 50.0 mL × 0.17 mol/L = 8.5 mmol

HCOONa: 80.0 mL × 0.12 mol/L = 9.6 mmol

Since HCOOH is a weak acid, it partially dissociates according to the equilibrium reaction:

HCOOH <=> HCOO− + H+

Given that the equilibrium constant (Ka) for HCOOH is 1.8 × 10−4, we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([A−]/[HA])

Where [A−] is the concentration of the conjugate base (HCOO−) and [HA] is the concentration of the weak acid (HCOOH). Plugging in the values:

pH = 3.74 + log(9.6 mmol/8.5 mmol) = 3.74 + log(1.13) = 3.74 + 0.05 = 3.79