Answer:
The volume of benzoic acid = 68.93 mL
The volume of benzoate = 31.07 mL
Explanation:
Since benzoic acid and sodium benzoate is a buffer of a weak acid/weak base, we can use the formula of Henderson-Hasselbalch equation.
pH = pKa + log [A-]/[HA]
with [A-] = Benzoate
with [HA] = Benzoic acid
This gives us:
4 = 4.2 + log [A-]/[HA]
[A-]/[HA] = 10^(4-4.2)
[A-]/[HA] = 10^-0.2
[A-]/[HA] = 0.631
This ratio is not the ratio of the initial concentration, but the ratio in the buffer ( after this buffer is finalized).
The solution has only one total volume, we only need to determine the initial volumes necessary to accomplish this 0.631 ratio.
0.631 = (0.140M *V(A-)) / (0.100M*V(HA)
Since the total voume = 100mL
V(A-) ≈ 100 mL - V(HA)
This gives us:
0.631 = (0.140M * (100 - VHA))/ (0.100M * VHA)
IF we say x = VHA we'll have the equation:
0.631 = 0.140 *(100- x)/(0.100 *x)
0.0631x +0.14x = 14
x = 68.93
V(HA) = 68.93 mL
This means V(A-) is 100 - 68.93 = 31.07 mL
We can control this by the following equation:
0.140 M benzoate * 31.07 mL / 100 mL = 0.0435M A-
0.100 M Benzoic acid * 68.93 mL /100 mL = 0.06893 M HA
[A-]/[HA] = 0.0435/0.06893 = 0.631
This means the volume of benzoic acid = 68.93 mL
The volume of benzoate = 31.07 mL
