Question

NOTE: I know the answer to this problem is 13.0 but I am unsure how to get the answer. A weak acid has a pKa of 7.3. 30 mL of 1.0 M NaOH is added to 100 mL of a 1.0 M buffer of this acid at pH 8.0. The resulting solution is at pH:

Answer 1

Solution.

The pH value of a buffer solution is determined by a Henderson-Hasselbach's equation:

$pH = pK_a+ log\frac{[Base]}{[Acid]};$

The initial buffer solution has a pH = 8,

$8 = 7.3+ log\frac{[Base]}{[Acid]};$

$\frac{[Base]}{[Acid]}=10^{8-7.3}=5.01;$

As the total concentration is 1.0 M,

$[Base]+[Acid]}=1.0;$

$[Base]=0.833M;$

$[Acid]=0.167M;$

The amounts of acid and base are (n=cV):

$[Acid]=0.0167mol;$

$[Base]=0.0833mol;$

The amount of a base added is

$[NaOH]=0.030\cdot 1.0 = 0.030mol;$

This amount of the base is enough to neutralize an acid completely, the amount of NaOH remaining is

n(NaOH)=0.03-0.0167=0.013 mol,

the total volume after the solutions has been mixed is 130 mL, the concentration of NaOH in solution is

$[OH^{-}]=\frac{0.013}{0.130}=0.1M;$