Question

How many grams of sodium acetate should be added into a two-liters buffer solution already containing 1 mole of acetic acid so as to prepare the solution with a final pH value of 4.0? The pKa of acetic acid is 4.75.
Furthermore, if you are required to adjust the aforementioned solution to a new pH value of 5.0, how many grams of NaOH should be added into the solution?

Answer 1

PH = PKa +log[salt]/[acid]

[CH3COOH] = mole/volume   = 1/2

[CH3COONa[   = moles/2

PH = PKa + log[CH3COONa]/[CH3COOH]

4    = 4.75+ logmole/1

4-4.75   =logmoles/1

-0.75   =logmole/1

moles   = 10^-0.75 = 0.1778 moles of CH3COONa

mass of CH3COONa = no of moles* molar mass

                                 = 0.1778*82 = 14.58gm of CH3COONa

By the addition of NaOH

no of moles of CH3COONa =0.1778+x

no of moles of CH3COOH    = 1-x

PH = PKa + log[CH3COONa]/[CH3COOH]

5    = 4.75+ log0.1778+x/1-x

5-4.75 = log0.1778+x/1-x

0.25   = log0.1778+x/1-x

0.1778+x/1-x = 10^0.25

0.1778+x   = 1.778(1-x)

0.1778+x-1.778+1.778x =0

-1.6002 = -2.778x

x = 0.576 moles

mass of NaOH = 0.576*40 = 23.04gm