Question

4. How much of (in mL) 0.15 M acetic acid (pKa 4.76) is needed to mix with 40 mL of 0.15 M sodium acetate to make a buffer of pH = 4.50 of 250 mL? (molar mass of acetic acid and sodium acetate are 60.0 g/mol and 82.0 g/ml, pKa of acetic acid is 4.76) (extra 10 pt)

Answer 1

4. ANSWER:

Concentration of acetic acid, [CH₃COOH] = 0.15 M

Volume of acetic acid, V_{acetic acid} = ?

Concentration of sodium acetate, [CH₃COONa] = 0.15 M

Volume of sodium acetate, V_{sodium 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 = pK_a - 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.