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A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared....

A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution.

The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places.

A buffered solution resists a change in pH.

Calculate the pH when 21.1 mL of 0.031 M HCl is added to 100.0 mL of the above buffer.

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Expert Solution

As per Henderson equation

pH = pKa + log [salt]/[acid]

pKa = -log ka = -log [ 1.8 x 10-5] = 4.74

[salt] = [sodium acetate] = 0.10

[acid] = [acetic acid] = 0.2

pH = 4.74 + log (0.1/0.2) = 4.43

on the addition of 0.031 M HCl, it will react with CH3COONa to form acetic acid and NaCl

100 mL of above buffer contains

0.01 moles of CH3COONa and 0.02 moles of acetic acid

(kindly note that molarity is the number of moles of solute in 1000 mL solution. Thus number of moles in 100 mL = molarity/10)

21.1 mL 0.031 M HCl contains (0.031/1000) x 21.1 = 0.0006541 moles

0.0006541 moles HCl reacts with 0.0006541 moles CH3COONa to form 0.0006541 moles acetic acid.

Thus total number of moles of acetic acid = 0.02 + 0.0006541 = 0.0206541 moles

total number of moles of sodium acetate = 0.01 - 0.0006541 = 0.0093459 moles

Total volume = 100 + 21.1 mL = 121.1 mL

New concentration of sodium acetate = (0.0093459/121.1) *1000 = 0.077 M

New concentration of acetic acid = (0.0206541/121.1)1000 = 0.1705

pH = pKa + log [salt]/[acid] = 4.74 + log [0.077/0.1705] = 4.39

queen_honey_blossom answered 2 years ago
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