Question

A buffer solution made from a generic acid (HA with Ka =4.82 x 10-5 )is 0.110 M in HA and 0.110 M in NaA. How many moles of NaOH can 200 mL of this buffer neutralize before the pH rises by 0.50?

Answer 1

0.634 mol of NaOH can 200 mL of this buffer neutralize before the pH rises by 0.50.

What is a buffer solution?

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components.

Solution:

A buffer consists of 0.11 M HA and 0.11 M NaA.

Ka =$4.82 X 10^-^5$

pKa = -log ($4.82 X 10^-^5$)

pKa = 11.83

Using Henderson equation to calculate pH of the buffer solution.

pH =PKa -log $\frac{Conjugate \;base}{Acid}$

pH = 11.83 - $\frac{0.11 M}{0.11 M}$

pH = 10.83

pOH = 14 - 10.83

pOH = 3.17

Calculating the moles of NaOH added by the addition of 200 mL

Molarity = $\frac{Moles \;of \;NaOH}{0.2 L}$

3.17 X 0.2 = Moles of NaOH

Moles of NaOH = 0.634 mol

Hence, 0.634 mol of NaOH can 200 mL of this buffer neutralize.

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