You need to prepare 100.0 ml of a ph=4.00 buffer solution using 0.100 m benzoic acid (pka = 4.20) and 0.140 m sodium benzoate. how much of each solution should be mixed to prepare this buffer?

ml benzoic acid = ?
ml sodium benzoate = ?

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

i believe the answer is true.