Question

You need to prepare an acetate buffer of pH 5.31 5.31 from a 0.784 M 0.784 M acetic acid solution and a 2.85 M KOH 2.85 M KOH solution. If you have 930 mL 930 mL of the acetic acid solution, how many milliliters of the KOH KOH solution do you need to add to make a buffer of pH 5.31 5.31 ? The p K a pKa of acetic acid is 4.76. 4.76.

Answer 1

Answer:

214 milliliters of KOH needs to be added in 1 litre of 0.784 M acetic acid to make a acetate buffer of 5.31

Explanation:

To solve the problem, let us first use the Henderson-Hasselbalch equation to determine the amount of acetate needed to make a buffer of pH 5.31.

Henderson-Hasselbalch equation:

$pH=pKa + log(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})$

Here, let us consider the moles of both species instead of the molar concentration, as the volume for both is the same. Also, acetate will be formed by the neutralization of acetic acid, hence the final moles of acetic acid will be the difference of initial moles of acetic acid and the moles of acetate formed. Now the equation becomes as follows:

$pH=pKa + log(\frac{n_{ CH_{3}COO^{-}}}{n_{iCH_{3}COOH}-n_{ CH_{3}COO^{-}}}})$

From given data

pH = 5.31

pKa = 4.76

n(CH₃COO⁻) = ?

ni(CH₃COOH) = 0.784 mol (initial moles of acetic acid)

Placing the data in the equation, we get:

$5.31=4.76 + log(\frac{n_{ CH_{3}COO^{-}}}{0.784-n_{ CH_{3}COO^{-}}}})\\ \\ n_{ CH_{3}COO^{-}}=10^{5.31-4.76}(0.784-(n_{ CH_{3}COO^{-}}))\\ \\ n_{ CH_{3}COO^{-}}= 2.78 mol-3.55(n_{ CH_{3}COO^{-}})\\ \\ n_{ CH_{3}COO^{-}}= 0.61mol$

The molar ratio of KOH and CH₃COOH is 1:1, i.e 1 mol of KOH will react with CH₃COOH and give 1 mol of acetate (CH₃COO⁻). Hence, 0.61 mol of KOH will give 0.61 mol of KOH. Now to determine the volume of 2.85 M KOH that contains 0.61 moles:

$M_{KOH} =\frac{n_{KOH} }{V_{KOH} (L)}$

$2.85=\frac{0.61}{V_{KOH} (L)}\\ \\ V_{KOH} (L)=\frac{0.61}{2.85}\\ \\ V_{KOH}=0.214 litre$

Finally convert liter into milliliter dividing by 1000 (mL/L)

Volume of KOH required = 214 milliliters