Question

Calculate the pH of a solution that contains 1.675 moles of benzoic acid and 0.925 moles of sodium benzoate (conjugate base of benzoic acid) in 8.00L. How many milliliters of 1.00 M NaOH must be added to raise the pH of the solution by 0.25?​

Answer 1

Apply

buffer equation; since there is conjugate and acid

pH = pKa + log(conjugate base / acid)

KA = 6.25*10^-5

pKa = -log(6.25*10^-5) = 4.20

pH = 4.20+ log(0.925/1.675) = 3.94212

pH = 3.94212

then

if we want to increase by 0.25

then

pH new = 3.94212 + 0.25 = 4.19212

substitute

pH = pKa + log(conjugate base / acid)

4.19212 = 4.20 + log(conjguate/acid)

approx

10^(4.19212 -4.20 ) = 0.982

conjugate = 0.982*acid

The changes of acid/base

base added = MV = 1*V

for the

conjugate = 0.925 + MV (incresess since more base, forms more conjugate)

acid = 1.675 -MV(decreasessince more base, decreases acid)

ehn

conjugate = 0.925 + 1*V

acid = 1.675 -1*V

and

conjugate = 0.982*acid

so

0.925 + 1*V = 0.982*(1.675 -1*V)

solve for V

0.925 +V = 0.982*1.675 -0.982*V

(1+0.982)V = 0.982*1.675-0.925

V = (0.982*1.675-0.925 )/((1+0.982)) = 0.3631937 L

V = 363 ml