Question

Question #b3509

Answer 1

!! LONG ANSWER !!

After the addition of the hydrochloric acid solution, the pH of the buffer will be equal to 3.97.

This time you're dealing with a buffer solution that consists of benzoic acid, `C_6H_5COOH`, which is a weak acid, and the benzoate anion, `C_6H_5COO^(-)`, which is its conjugate base.

When you add the hydrochloric acid solution, the strong acid will react with the conjugate base to produce more benzoic acid. In the process, the hydrochloric acid will be completely consumed.

The number of moles of benzoic acid and of benzoate anions initially present in the solution are

`C = n/V => n = C * V`

`n_"benzoic acid" = "0.150 M" * "1.00 L" = "0.150 moles"`

`n_"benzoate" = "0.250 M" * "1.00 L" = "0.250 moles"`

The number of moles of hydrochloric acid you add to the buffer is equal to

`n_(HCl) = "1.00 M" * 100 * 10^(-3)"L" = "0.100 moles"`

You can use an ICF table to help you manage the number of moles of each compound that get consumed/produced. The balanced chemical equation for this reaction looks like this

`" "C_6H_5COO_((aq))^(-) + HCl_((aq)) -> C_6H_5COOH_((aq)) + Cl_((aq))^(-)`
I..........0.250......................0.100...................0.150
C.......(-0.100)....................(-0.100)................(+0.100)
F.........0.150..........................0.........................0.250

You started with 0.250 moles of benzoate ions and 0.150 moles of benzoic acid, and ended up with 0.150 moles of benzoate ions and 0.250 moles of benzoic acid.

All the hydrochloric acid was consumed by the reaction.

The total volume of the solution will be

`V_"total" = V_"initial" + V_(HCl)`

`V_"total" = "1.00 L" + 100 * 10^(-3)"L" = "1.10 L"`

The new concentrations of the benzoic acid and of the benzoate ions will be

`[C_6H5COOH] = "0.250 moles"/"1.10 L" = "0.2273 M"`

`[C_6H_5COO^(-)] = "0.150 moles"/"1.10 L" = "0.1364 M"`

The `pK_a` of the acid is

`pK_a = -log(K_a) = -log(6.5 * 10^(-5)) = 4.19`

Use the Henderson-Hasselbalch equation to determine the pH of the solution

`pH_"sol" = pK_a + log(([C_6H_5COO^(-)])/([C_6H_5COOH]))`

`pH_"sol" = 4.19 + log((0.1364cancel("M"))/(0.2273cancel("M"))) = color(green)(3.97)`

SIDE NOTE The initial pH of the buffer was

`pH_"initial" = 4.19 + log(0.250/0.150) = 4.41`