100 mL of 0.2 M weak acid HA is mixed with 50 mL of 0.2 M NaOH solution at 298 K. If the base dissociation constant ( K_b ) for the conjugate base A⁻ is 10⁻⁹ , what is the pH of the resulting mixture?

Step-by-step solution

First, calculate the initial millimoles of the weak acid HA and the strong base NaOH : Millimoles of HA = 100 mL 0.2 M = 20 mmol Millimoles of NaOH = 50 mL 0.2 M = 10 mmol The neutralization reaction is: HA + NaOH NaA + H₂O Since NaOH is the limiting reagent, 10 mmol of NaOH will react with 10 mmol of HA to form 10 mmol of the salt NaA . After the reaction, the mixture contains: Remaining HA = 20 - 10 = 10 mmol Formed A⁻ = 10 mmol This forms an acidic buffer where the concentration of the weak acid equals the conce