Question

In: Chemistry

3. a) Calculate the pH of a sodium acetate-acetic acid buffer solution (Ka= 1.78 x 10-5)...

3. a) Calculate the pH of a sodium acetate-acetic acid buffer solution (Ka= 1.78 x 10-5) in
which the concentration of both components is 1.0 M. What would be the pH of the
solution if 1.5 mL of 0.50 M NaOH was added to 35.0 mL of the buffer? How much
does the pH change?

b) If the same amount and concentration of NaOH was added to 35.0 mL of pure water
(initial pH = 7), what would be the resulting pH change?

Solutions

Expert Solution

3)

a) pH of acidic buffer = pka + log(ch3coona+NaOH/ch3cooh-NaOH)

pka of ch3cooh = -logKa = -log(1.78*10^-5) = 4.75

No of mole of acetic acid present in buffer = 35*1 = 35 mmole

No of mole of sodium acetate = 35*1 = 35 mmol

No of mole of NaOH added = 1.5*0.5 = 0.75 mmole

pH = 4.75 + log((35+0.75)/(35-0.75))

   = 4.77

pH change = 4.77-4.75 = 0.02

b)

Concentration of NaOH solution = n/v

                               = 1.5*0.5/(35+1.5)

                               = 0.0205 M
pOH = - log[OH-]

    = -log0.0205

    = 1.7

pH = 14-pOH = 14-1.7 = 12.3

initial pH of water = 7

pH change = 12.3 - 7 = 5.3


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