Question

How much 5.90 M NaOH must be added to 610.0 mL of a buffer that is .0210 M acetic acid and .0245 M sodium acetate to raise the pH to 5.75?

Answer 1

initially:

pH = pKa + loG(Acetate/cid)

5.75 = 4.75 + log(Acetate/cid)

Acetate/Acid = 10^(5.75-4.75) = 10

then

Acetate/Acid = 10

mmol of acid = M1V1 = 610*0.021 = 12.81 mmol

mmol of acetate = M2V2 = 610*0.0245 = 14.945 mmol

after adding:

Mbase*Vbase = 5.90*Vbase

then

mmol of acid = 12.81 - 5.9Vbase

mmol of acetate = 14.945 + 5.9Vbase

and we know that

Acetate/Acid = 10

substitute both equations:

(14.945 + 5.9Vbase)/(12.81 - 5.9Vbase) = 10

(14.945 + 5.9Vbase) = (128.1 - 59Vbase)

14.945 -128.1 = -(-59-5.9)Vbase

Vbas = (14.945 -128.1)/(-59-5.9) = 1.74 ml