Question

you want to make 100ml of 0.20M acetic acid buffer with pH=50. you are given a stock of 1.0M acrtic acid and a bottle of solid sodium acetate sale (MW=82g/mol).
what is the ratio of [A-]/[HA] when your buffer pH is 5.0?
determine the concentration of weak acid and conjugate base in the buffer.
how many mL of 1.0 acetic acid will you need?
how many grams of sodium acetate will you need?
if you now have add 1mL of 1M NaOH to your buffer, what would be the resulting pH?

Answer 1

what is the ratio of [A-]/[HA] when your buffer pH is 5.0?

this can be calcualated using Henderson Hasselback equations

pH = pKa + log(A-/HA)

if pH = 5, and pKa for acetic acid = 4.75

then

5.0= 4.75+ log(A-/HA)

[A-]/[HA] = 10^(5.0-4.75)

[A-]/[HA] =1.7782

determine the concentration of weak acid and conjugate base in the buffer.

given

[A-]/[HA] =1.7782

total molarity = MV = 100*0.2 = 20

mmol of A- + mmol of HA = 20 mmol

let

[A-]/[HA] =1.7782 --> mmol of A- / mmol of HA = 1.7782

mmol of A- = 1.7782 *mmol of HA

now, substitute

mmol of A- + mmol of HA = 20 mmol

1.7782 *mmol of HA + mmol of HA = 20 mmol

mmol of HA = 20/(1+1.7782 ) = 7.19

mmol of A-= 20-7.19 = 12.81 mmol

since V = 100 mL

[HA] = mmol/mL = 7.19/100 = 0.0719 M

[A-] = 12.81 /100 = 0.1281 M

how many mL of 1.0 acetic acid will you need?

mmol of HA required = 7.19

[HA] = mmol/mL

mL = mmol/[HA] = 7.19 / 1 = 7.19 mL of acetic acid stock solution required

how many grams of sodium acetate will you need?

mmol of A- = mmol of NaA = 12.81*82 = 1050.42 mg = 1.05 g of sodium acetate

if you now have add 1mL of 1M NaOH to your buffer, what would be the resulting pH?

mmol of OH- = MV = 1*1 = 1

mmol of HA left = 7.19-1 = 6.19

mmol of A- formed = 12.81+1 = 13.81

substitute

pH = 4.75+ log(A-/HA)

pH = 4.75+ log(13.81/6.19) = 5.098

pH change + 0.098 units