# Question aa487

Aug 9, 2015

Alternatively, you could use the Henderson-Hasselbalch equation.

#### Explanation:

Since you're dealing with a buffer, which in your case is a solution that contains a weak acid, acetic acid, and its conjugate base, the acetate anion, in comparable amounts, you can use the Henderson-Hasselbalch equation to determine its pH.

pH_"sol" = pK_a + log ( (["conjugate base"])/(["weak acid"]))

Here $p {K}_{a}$ is equal to

$p {K}_{a} = - \log \left({K}_{a}\right)$

$p {K}_{a} = - \log \left(1.76 \cdot {10}^{- 5}\right) = 4.75$

Use the classic approach to determine the concentrations of the weak acid and of the conjugate base after the sodium hydroxide solution is added.

Since you add together $1 \cdot {10}^{- 3} \text{moles}$ of sodium hydroxide and $3 \cdot {10}^{- 3} \text{moles}$ of acetic acid, you will be left with

${n}_{\text{acetic acid" = 3 * 10^(-3) - 1 * 10^(-3) = 2 * 10^(-3)"moles}}$

The reaction will produce the same number of moles of acetate ions as the number of moles of sodium hydroxide consumed. This means that you will get

[CH_3COOH] = (2 * 10^(-3)"moles")/(40 * 10^(-3)"L") = "0.05 M"

[CH_3COO^(-)] = (1 * `10^(-3)"moles")/(40 * 10^(-3)"L") = "0.025 M"

Therefore, the solution's pH is

pH_"sol" = 4.75 + log ((0.025cancel("M"))/(0.05cancel("M")))#

$p {H}_{\text{sol}} = 4.75 + \left(- 0.301\right) = \textcolor{g r e e n}{4.45}$