Question

For this excercise you will prepare two buffer systems called Buffer A1 and Buffer P1 with a target pH value of 7. You will test the ability of these two buffers to resist pH change adding a fixed amount of sodium hydroxide solution to each and measuring and observing the pH change. Buffer A1 will be composed of 0.1M acetate buffer by combining a stock solutions of acetic acid and sodium acetate. Buffer P1 will be 0.1M phosphate buffer prepared by combining stock solutions of diabasic sodium phosphate and monobasic sodium phophate.

1. If the total buffer concentration ([acetic acid] + [acetate]) is 0.1M, what is the molar concentration of sodium acetate at pH 7?

2. How many grams of sodium acetate are needed in 50mL of a pH 7 0.1M buffer?

please show steps.

Answer 1

Buffer A1 is buffer composed of weak acid (acetic acid) and salt of weak acid (sodium acetate). pH of the buffer is given by Henderson Hasselbach equation,

pH = pK_a + log { [salt] / [acid] }

pH is negative logarithm of concentration of H⁺ ions in buffer solution; pKa is negative logarithm of acid dissociation constant (K_a) of the weak acid; [acid] and [salt] are concentrations of the weak acid and its salt respectively.

1. Molar concentration of sodium acetate

pK_a of acetic acid = 4.75;

{ For acetic acid, K_a = 1.76 * 10^-5; pK_a = -logK_a = log(1/K_a) = log [ 1 / (1.76 * 10^-5)] = 4.75 }

given pH =7;

So for buffer A1, Henderson-Hasselbalch equation is,

pH = pKa + log { [acetate] / [acetic acid] }

7 = 4.75 + log { [acetate] / [acetic acid] }

log { [acetate] / [acetic acid] }​ = 7-4.75 = 2.25

{ [acetate] / [acetic acid] }​ = antilog 2.25 = 177.83

Therefore, { [acetate] / [acetic acid] } = 177.83

or [acetic acid] = [acetate] * (1/177.83)

[acetic acid] = [acetate] * (0.005623)

Given, [acetate] + [acetic acid] = 0.1

So, [acetate] + [acetate]*(0.005623) = 0.1 or 1.005623 * [acetate] = 0.1

[acetate] = 0.1 / 1.005623 = 9.944 * 10^-2 M

Molar concentration of sodium acetate in the buffer at pH 7, is, 9.944 * 10^-2 M

2. Grams of sodium acetate needed in 50mL of pH 7.01 M buffer

From answer to question1, we have,

[acetate] = 9.944 * 10^-2 M

So, [acetate] = 9.944 * 10^-2 mol / L (since M is mol / L)

So, 1 L or 1000 mL of the buffer solution contains 9.944 * 10^-2 moles of sodium acetate.

Hence, 50 mL of the buffer solution contains, 50 * (9.944 * 10^-2) / 1000 = 4.972 * 10^-3 mol

50 mL of the buffer solution contains, 4.972 * 10^-3 mol of sodium acetate

Molar mass of sodium acetate = 82.0343 g / mol

mass = (number of moles) * (molar mass)

mass of sodium acetate = (4.972 * 10^-3) * (82.0343) = 0.4079 g

So, 0.4079 g of sodium acetate are needed in 50 mL of a pH7 0.1M buffer.

ANSWERS: 1. 9.944 * 10^-2 M

   2. 0.4079 g