Question

Calculate the pH of each of the following 8 solutions, the first four made by adding HCl to a 0.1 M solution of sodium acetate, and the second four made by adding the same amounts of HCl to a beginning solution of 0.01M sodium acetate. The amounts of HCl added were 0.0025 M, 0.005 M, 0.01 M or 0.05 M of the strong acid.

Answer 1

HCl + CH3COONa ----> CH3COOH + NaCl

i) 0.0025M HCl and 0.1M CH3COONa

After adding HCl-

CH3COONa = 0.1M - 0.0025M = 0.0975M

CH3COOH = 0.0025M

Ka for CH3COOH = 1.8*10^-5

Using henderson hasslebach equation,

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = -log[1.8*10^-5] + log[0.0975M/0.0025M]

pH = 4.74 + 1.591

pH = 6.33

(ii) 0.005M HCl + 0.1M CH3COONa

After adding HCl-

CH3COONa = 0.1M-0.005M = 0.095M

CH3COOH = 0.005M

pH = pKa + log[0.095/0.005]

pH = 4.74 + 1.28

pH = 6.02

(iii) 0.01M HCl + 0.1M CH3COONa

After HCl-

CH3COONa = 0.1M-0.01M = 0.09M

CH3COOH = 0.01M

pH = 4.74 + log[0.09/0.01]

pH = 4.74 + 0.95

pH =5.69

(iv) 0.05M HCl + 0.1M CH3COONa

After adding HCl

CH3COONa = 0.1M-0.05M = 0.05M

CH3COOH = 0.05M

pH = 4.74 + log[0.05M/0.05M]

pH = 4.74

(v) 0.0025M HCl + 0.01M CH3COONa

After adding HCl-

CH3COONa = 0.01M-0.0025M = 0.0075M

CH3COOH = 0.0025M

pH = 4.74+ log[0.0075M/0.0025M]

pH = 4.74+ 0.48

pH = 5.22

(vi) 0.005M HCl + 0.01M CH3COONa

After adding HCl

CH3COONa = 0.01M-0.005M = 0.005M

CH3COOH = 0.005M

pH = 4.74 + log[0.005M/0.005M]

pH = 4.74

(vii) 0.01M HCl + 0.01M CH3COONa

After HCl-

CH3COONa = 0.01M-0.01M = 0

CH3COOH = 0.01M

pH =4.74+ log[0/0.01]

pH = 4.74

(viii) 0.05M HCl + 0.01M CH3COONa

After addition-

HCl = 0.05M- 0.01M = 0.04M

CH3COONa = 0.01M-0.01M =0

CH3COOH = 0.01M

pH = 4.74