Question

How much 5.60 M NaOH must be added to 540.0 mL of a buffer that is 0.0205 M acetic acid and 0.0270 M sodium acetate to raise the pH to 5.75?

Answer 1

this will form a buffer so

pH = pKa+ log(acetate/acetic adi)

pKa = 4.75

solve for ratio

pH = pKa+ log(acetate/acetic adi)

5.75 = 4.75+ log(acetate/acetic adi)

(acetate/acetic acid) = 10^(5.75-4.75) = 10

acetate = (acetic acid)*10

mmol of acid = MV1 - 5.6*V3 = 540*0.0205 -5.6*V3

mmol of acetate = MV1 + 5.6V3 = 540*0.027 + 5.6V3

substitute

acetate = (acetic acid)*10

540*0.027 + 5.6V3 = 10(540*0.0205 -5.6*V3)

540*0.027 - 10*540*0.0205 = 5.6V3 - 56V3

V3 = (540*0.027 - 10*540*0.0205)/(5.6-56)

V3 = 1.907 ml of base needed