Question

How much 5.80 M NaOH must be added to 500m0 mL of a buffer that is .0190 M acetic acid and .0265 M sodium acetate to raise the pH to 5.75?

Answer 1

M = 5.8 M of NaOH

V = 500 ml buffer

M = 0.019 HAc

M = 00265 M of NaAc

pH = 5.75

the buffer equation:

pH = pKa + log(NaAc / HAc)

pKa = 4.75 (from tables)

original pH

pH = 4.75 + log(0.0265/0.019) = 4.89449

if we want

5.75 then

5.75 = 4.75+ log(NaAc / HAc)

ratio = 10^(5.75-4.75) = (NaAc / HAc)

(NaAc / HAc) = 10

that is

NaAc = HAc*10 (Equatio n1)

note that when we add NaOh

this hpapens

HA + NaOH = H2O + NaAc

there is: DECREASE in acid, INCREASE in CONJUGATE

so

[NaAc] = Original + added base

[HAc] = original - added base

[NaAc] = (0.0265 + 5.8V)/(500 + V)

[HAc] =  (0.019- 5.8V)/(500 + V)

From equation 10[HAc] = [NaAc] then

[NaAc] = (0.0265 + 5.8V)/(500 + V) turns to:

10[HAc] =  (0.0265 + 5.8V)/(500 + V)

and we have also

[HAc] =  (0.019- 5.8V)/(500 + V)

therefore

[HAc] =  (0.019 - 5.8V)/(500 + V)

10[HAc] =  (0.0265 + 5.8V)/(500 + V)

solve simultaneously for V and HAc

(1) in (2) gives

10*( (0.019- 5.8V)/(500 + V)) = (0.0265 + 5.8V)/(500 + V)

0.19 -58V / (500 + V) = (0.0265 + 5.8V)/(500 + V)

0.19 -58V= (0.0265 + 5.8V)

V*(-58-5.8) = 0.0265 -0.19

V = -( 0.0265 -0.19)/(58-5.8) = 0.0031321839 Liter or 3.13 ml

Then

add 3.13 ml of Base