Question

500.0 mL of 0.150 M NaOH is added to 555 mL of 0.200 M weak acid (Ka = 6.24 × 10-5). What is the pH of the resulting buffer?

Answer 1

The Acid -Base equilibria is given by-

From the above amounts of number of moles in the solution,we see that the NaOH is in lesser amount present in the solution.So,NaOH is the Limiting Reagent here i.e. NaOH would be consumed entirely in the reaction.

At Equilibrium,the No. of moles of weak acid and the No of moles of its Conjugate Base will be-

now,using the Henderson equation for Buffers,the pH of the solution is given by-

Hence the pH of the resulting buffer is-4.524.