50 mL of 0.1 M CH 3 COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be _____________ 10 -2 . (Nearest integer) (Given : pK a (CH 3 COOH) = 4.76)<

CH 3 COOH + NaOH CH 3 COONa + H 2 O After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml Initially, Number of millimole of NaOH = 25 0.1 = 2.5 mm Number of millimole of CH 3 COOH = 50 0.1 = 5 mm After nutrilisation, Millimole of NaOH = 0 Millimole of C

50 mL of 01 M CH3COOH is being titrated against 01 M NaOH Wh: null (Chemistry)

CH₃COOH + NaOH

\to

CH₃COONa + H₂O

After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml

Initially,

Number of millimole of NaOH = 25

\times

0.1 = 2.5 mm

Number of millimole of CH₃COOH = 50

\times

0.1 = 5 mm

After nutrilisation,

Millimole of NaOH = 0

Millimole of CH₃COOH = 5

-

2.5 = 2.5 mm

Millimole of CH₃COONa = 2.5

After nutrilisation,

Concentration of CH₃COOH =

[C{H_3}COOH] = {{5 - 2.5} \over {75}} = {1 \over {30}}

Concentration of CH₃COONa =

[C{H_3}COONa] = {{ 2.5} \over {75}} = {1 \over {30}}

{P^H} = {P^{Ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}

= 4.76 + \log {{{1 \over {30}}} \over {{1 \over {30}}}}

= 4.76 + \log (1)

= 4.76 + 0

= 4.76

= 4.76 \times {10^{ - 2}}

Explanation

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