Question

Question #aa487

Answer 1

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 `pK_a` is equal to

`pK_a = -log(K_a)`

`pK_a = -log(1.76 * 10^(-5)) = 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 * 10^(-3)"moles"` of sodium hydroxide and `3 * 10^(-3)"moles"` of acetic acid, you will be left with

`n_"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")))`

`pH_"sol" = 4.75 + (-0.301) = color(green)(4.45)`