Question

8. You have prepared a 0.2 M acetate solution with a pH = 4.76. Determine the resulting pH when the acids and bases below are added to the solution. The pKa of CH3COOH is 4.76.

a. 0.025 M HCl

b. 0.050 M HCl

c. 0.025 M NaOH

d. 0.050 M NaOH

e. 0.050 M HCl and 0.025 M NaOH

f. 0.025 M HCl and 0.050 M NaOH

Answer 1

Note: Since volume is not given in the question, i have calculated by considering equal volume of both the acetate solution and the acid/base i.e if we add 100 mL of 0.025 M HCl to 100 mL of 0.2M acetate buffer solution

8. Given the pH of the acetate buffer is equal to the pKa of acetic acid. Hence according to Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> 4.76 = 4.76 +  log[CH3COO-] / [CH3COOH]

=> [CH3COO-] / [CH3COOH] = 1

=>  [CH3COO-] = [CH3COOH]

Since the concentration of the acetate solution is 0.2 M

[CH3COO-] = [CH3COOH] = 0.1 M

(a): When 0.025 M HCl is added it will react with CH3COO- to form CH3COOH.

0.025 mol of HCl will react with 0.025 mol of CH3COO-, if equal volume of both are taken. Hence concentration of CH3COO- and CH3COOH after reaction is

[CH3COO-] = 0.075 / Vt M , [CH3COOH] = 0.125 / Vt M

Now applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> pH = 4.76 + log[(0.075/Vt) / (0.125/Vt)] = 4.76 + log(0.075 / 0.125) = 4.54

(b): When 0.050 M HCl is added it will react with CH3COO- to form CH3COOH.

0.050 mol of HCl will react with 0.050 mol of CH3COO-, if equal volume of both are taken. Hence concentration of CH3COO- and CH3COOH after reaction is

[CH3COO-] = 0.050 / Vt M , [CH3COOH] = 0.150 / Vt M

Now applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> pH = 4.76 + log[(0.050/Vt) / (0.150/Vt)] = 4.76 + log(0.050 / 0.150) = 4.28

(c): When 0.025 M NaOH is added it will react with CH3COOH to form CH3COO-.

0.025 mol of NaOH will react with 0.025 mol of CH3COOH, if equal volume of both are taken. Hence concentration of CH3COO- and CH3COOH after reaction is

[CH3COO-] = 0.125 / Vt M , [CH3COOH] = 0.075 / Vt M

Now applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> pH = 4.76 + log[(0.125/Vt) / (0.075/Vt)] = 4.76 + log(0.125/0.075) = 4.98

(d): When 0.050 M NaOH is added it will react with CH3COOH to form CH3COO-.

0.050 mol of NaOH will react with 0.050 mol of CH3COOH, if equal volume of both are taken. Hence concentration of CH3COO- and CH3COOH after reaction is

[CH3COO-] = 0.150 / Vt M , [CH3COOH] = 0.050 / Vt M

Now applying Hendersen equation

pH = pKa + log[CH3COO-] / [CH3COOH]

=> pH = 4.76 + log[(0.150/Vt) / (0.050/Vt)] = 4.76 + log(0.150/0.050) = 5.24