In: Chemistry
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?
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