4. ANSWER:
Concentration of acetic acid, [CH3COOH] = 0.15 M
Volume of acetic acid, Vacetic acid = ?
Concentration of sodium acetate, [CH3COONa] = 0.15 M
Volume of sodium acetate, Vsodium acetate = 40 mL
pH = 4.50
Number of moles of sodium acetate = volume x concentration = 40 mL x 0.15 M = 6.0 x 10-3 moles
Now,
according to Handerson-Hasselbaltch equation:
pH = pKa - log{[# of moles of acetic acid]/[# of moles of sodium acetate]}
4.50 = 4.76 - log{[# of moles of acetic acid]/[6.0 x 10-3 mol]}
log{[# of moles of acetic acid]/[6.0 x 10-3 mol]} = 0.26
{[# of moles of acetic acid]/[6.0 x 10-3 mol]} = antilog{0.26} = 1.82
# of moles of acetic acid = 1.82 x 6.0 x 10-3 mol = 0.011 moles
And,
Volume of acetic acid = {# of moles of acetic acid} / {concentration}
= {0.011 mol} / {0.15 M}
= 0.07333 L
= 73.33 mL
Hence, 73.33 mL of 0.15 M acetic acid is needed to mix with 40 mL of 0.15 M sodium acetate to make buffer of pH = 4.50.
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