Answer to Question #349285 in General Chemistry for Mellisa

Question

A 400 mL buffer solution was prepared by reacting sodium acetate, CH3COONa,

with 1.00 M acetic acid, CH3COOH, with a pH of 4.56 Calculate the mass of

CH3COONa required to prepare the solution. Ka CH3COOH= 1.75 x 10-5

Answer 1

1.

Using pH, we find the concentration of H + ions

pH = 4,56

pH= -log [H⁺]

4,56 = -log [H⁺]

[H⁺] = 2,76 * 10^-5

2.

Using the H + ion concentration and the Ka value, we find the CH3COOH concentration. for this we divide the H + ion concentration by K_a:

CH₃COOH => CH₃COO^– + H⁺

[CH₃COOH] = [H⁺] /K_a =2,76*10^-5/1,75*10^-5 =

= 1,577 mol/l

3.

CH₃COONa + H₂O = CH₃COOH + NaOH

We find the amount of CH₃COONa for this we separate the CH₃COOH concentration given in the task from the CH₃COOH concentration found Then multiply it by 0.4 l

1,577 M - 1 M = 0,577 M

n= C_M * V = 0,577 * 0,4 = 0,2308 mol

4.

M_r (CH₃COONa) = 82 g/mol

We find the mass of CH₃COONa using the following formula:

m = n * M_r = 0,2308 * 82 = 18,9 g

Answer: 18,9 g CH₃COONa