Question

7. Calculate the mass of sodium acetate (CH3COONa) that must be added to 1.00 L 0.450 M acetic acid (CH3COOH), Ka = 1.8 x 10 ) to form a pH = 5.00 buffer. Ka = 1.8 x 109.

Answer 1

Given

Overall volume of solution = 1 L

Concentration of CH₃COOH = 0.450M

Required pH of solution = 5

K_a(CH₃COOH) = 1.8 x 10^-5

Molar mass of sodium acetate = 82.03 g/mol

We need to calculate weght of sodiun acetate that has to be added.

$CH_3COOH\: \leftrightharpoons \: CH_3COO^-+H^+$

$K_a(CH_3COOH)\, = \: \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

$\Rightarrow [CH_3COO^-]\: = \: \frac{K_a [CH_3COOH]}{[H^+]}\: \: \: \:\: \: \: \: -------(1)$

$pH\: =\: -log[H^+]$

$\Rightarrow 5\: =\: -log[H^+]$

$\Rightarrow [H^+]\: =\: 10^-^5$

$\therefore\: From\: \: (1)$

$\Rightarrow [CH_3COO^-]\: = \: \frac{1.8\times 10^-^5\times 0.450}{[10^-^5]}$

$\Rightarrow [CH_3COO^-]\: = \: 0.81\: M$

So required concentration of $[CH_3COO^-]\:\: is\: \: 0.81\: M$ .

$Now, \: \: \: \: \frac{n}{V}\:= \: 0.81\: \: \: \: \: \: \: \: \: \: \: \: \: where \: \: n\: \: is \: \: moles \: \: of \: \: CH_3COO^-$

$\Rightarrow \: \: \: \: \frac{\frac{mass}{Molar\: \: \: \: mass}}{V}\:= \: 0.81\:$

$\Rightarrow \: \: \: \: \frac{\frac{mass}{82.03}}{1}\:= \: 0.81\:$

$\Rightarrow \: \: \: \:mass\: = \: 0.81\times 82.03$

$\Rightarrow \: \: \: \:mass\: = \: 66.4443\, g$

So 66.4443 g of Sodium acetate has to be added