Question

How much 5.90 M NaOH must be added to 530.0 ml of a buffer that is 0.0200 M acetic acid and 0.0255 M sodium acetate to raise the pH to 5.75?

Answer 1

How much 5.90 M NaOH must be added to 530.0 ml of a buffer that is 0.0200 M acetic acid and 0.0255 M sodium acetate to raise the pH to 5.75?

pH = pKa + log ([A-] / [HA])

Here A- = the conjugate base of the buffer which comes from sodium acetate

HA = the weak acid of the buffer means acetic acid

The value of pKa for acetic acid is 4.74.

Now calculate the moles of acid and base:

initial moles HA = molarity * volume in L

= (0.0200)(0.530)

= 0.0106 moles HA
initial moles A- = = molarity * volume in L

= (0.0255)(0.530)

= 0.0135 moles A-

Here pH = 5.75 now we calculate the ratio of A- / HA as follows:

5.75 = 4.74 + log (moles A- / moles HA)

log (moles A- / moles HA) = 0.96

(moles A- / moles HA) =10^0.96

moles A- / moles HA = 9.12

When we added NaOH it convert some HA to A- by reacting it with OH- as follows:

NaOH < == > Na+   OH-
HA + OH- ==> H2O + A-

moles of HA + moles A- = 0.0106 + 0.0135 = 0.0241

moles A- = 0.0241- moles HA

moles A- / moles HA = 9.12

now put this moles A- = 0.0241- moles HA value in the following equation:

moles A- / moles HA = 9.12

0.0241- moles HA / moles HA = 9.12

0.0241 - moles HA = (9.12)(moles HA)
0.0241 = 10.12 moles HA
moles HA = 2.38*10^-3 or 0.00238

Now we calculate the moles of NaOH which reduce the moles of acid from HA from 0.0106 to 0.00238.

Since HA reacts with OH- in a 1:1 ratio, then moles OH- needed

= 0.0106 - 0.00238= 0.00822 moles OH-
0.00822 moles OH- = 0.00822 moles NaOH

Now calculate the volume of NaOH :

Molarity = number of mole s/ volume in L

Volume in L = number of moles / molarity

= 0.00822 moles NaOH/5.90 M

=1.39*10^-3 L or 1.39 ml