Question

calculate the ph of acidic buffer containing 0.1 M CH3COOH and 0.5 CH3COONa(ka for CH3COOH is 1.8×10^-6 )?​

Answer 1

Explanation:

For 0.1 M CH

3

COOH & 0.1 CH

3

COONa

pH=pKa+log

0.1

0.1

=Pka

PKa=4.74−(1)

When 0.05 moles of HCl is added to 1Lit of the solution

CH

3

COONa+HCl⟶CH

3

COOH+NaCl

Initially 0.1 M0.050.1 M

1lit 1lit

0.1 mole

Finally 0.05 mole 00.15 mole

PH

Now

=PKa+log

[acid]

[salt]

=4.74+log

0.15

0.05

=4.74−log3=4.27