We need 1.00 L of an acetic acid/acetate (K a = 1.80x10 -5 ) buffer with a start

Question

We need 1.00 L of an acetic acid/acetate (Ka= 1.80x10-5) buffer with a starting pH of 5.000. 500.0 mL of the buffer must be able to resist the addition of 7.50 mL of 0.250 M NaOH while remaining below pH 5.250. How many mL of 0.500 M stock acid and 0.250 M NaOH base solution must be used in preparation of this buffer? Check both answers.

22.4 mL acid

22.4 mL base

31.5 mL acid

31.5 mL base

40.5 mL acid

40.5 mL base

45.0 mL acid

45.0 mL base

63.0 mL acid

63.0 mL base

72.0 mL acid

72.0 mL base

81.0 mL acid

81.0 mL base

126 mL acid

126 mL base

144 mL acid

144 mL base

311 mL acid

311 mL base

Explanation / Answer

According to Henderson-Hasselbulch equation, pH = pKa + Log(nCH3COONa +nnaOH/nCH3COOH -nNaOH), where n = no. of mmol

a) The volume of acid = 22.4 mL, i.e. the no. of mmol of CH3COOH = 0.5 mmol/mL * 22.4 mL = 11.2 mmol

The volume of base = 22.4 mL, i.e. the no. of mmol of NaOH = 0.25 mmol/mL * 22.4 mL = 5.6 mmol

Here, 5.6 mmol of NaOH reacts with 5.6 mmol of CH3COOH to form 5.6 mmol of CH3COONa leaving 11.2-5.6 = 5.6 mmol of CH3COOH

The no. of mmol of added base to buffer solution = 7.5 mL * 0.25 mmol/mL = 1.875 mmol

i.e. pH = 4.74 + Log(5.6+1.875/5.6-1.875) = 5.04

(ii) The volume of acid = 31.5 mL, i.e. the no. of mmol of CH3COOH = 0.5 mmol/mL * 31.5 mL = 15.75 mmol

The volume of base = 31.5 mL, i.e. the no. of mmol of NaOH = 0.25 mmol/mL * 31.5 mL = 7.875 mmol

Here, 7.875 mmol of NaOH reacts with 7.875 mmol of CH3COOH to form 7.875 mmol of CH3COONa leaving 15.75-7.875 = 7.875 mmol of CH3COOH

i.e. pH = 4.74 + Log(7.875+1.875/7.875-1.875) = 4.95

(iii) The volume of acid = 40.5 mL, i.e. the no. of mmol of CH3COOH = 0.5 mmol/mL * 40.5 mL = 20.25 mmol

The volume of base = 40.5 mL, i.e. the no. of mmol of NaOH = 0.25 mmol/mL * 40.5 mL = 10.125 mmol

Here, 10.125 mmol of NaOH reacts with 10.125 mmol of CH3COOH to form 10.125 mmol of CH3COONa leaving 20.25-10.125 = 10.125 mmol of CH3COOH

i.e. pH = 4.74 + Log(10.125+1.875/10.125-1.875) = 4.90

Here, the pH value keep on decreasing.

Hence, most likely the volumes of 22.4 mL of acid and 22.4 mL of the base will give the required pH of ~ 5.0

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