Question

1713

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150M NaOH solution. Calculate the pH after the following volumes of base have been added.

a) 0 mL

b) 17.5 mL

c) 34.5 mL

d) 35.0 mL

e) 35.5 mL

f) 50.0 mL

Answer 1

a) Before addition of Base

pH of weak acid = 1/2(pka-log C)

pka of CH3COOH = 4.74

c = CONCENTRATION OF ch3cooh = 0.15 M

pH = 1/2(4.74-log0.15)

   = 2.8

b) after addition of 17.5 ml of NaOH

No of mol of NaOH = 17.5/1000*0.15 = 0.002625 mol

No of mol of CH3COOH = 35.0/1000*0.15 = 0.00525 mol

pH = pka + log(base/acid)

= 4.74 + log(0.002625/(0.00525-0.002625))

= 4.74

c) after addition of 34.5 ml of NaOH

No of mol of NaOH = 34.5/1000*0.15 = 0.005175 mol

No of mol of CH3COOH = 35.0/1000*0.15 = 0.00525 mol

pH = pka + log(base/acid)

= 4.74 + log(0.005175/(0.00525-0.005175))

= 6.6
d)
at equivalence point
pH depeds upon only base

Concentration of base = 35/70*0.15 = 0.075 M

pH of salt = 7+1/2(pka+logC)

        = 7+1/2(4.74+log0.075)

    = 8.8
e) ph depends upon excess vase added

concentration of excess base = 0.5/70.5*0.15 = 0.0011 M

pOH = -log(OH-) = -log0.0011 = 3

pH = 14-3 = 11

f) ph depends upon excess vase added

concentration of excess base = 15/85*0.15 = 0.0265 M

pOH = -log(OH-) = -log0.0265 = 1.58

pH = 14-1.58 = 12.42