Question

a liter of a solution containing 0.1 mole of CH3COOH and 0.1 mole of CH3COONa provide buffer of pH 4.74 calculate thepH of the solution after the addition of 0.02 mol of NaoH ka=1.8×10^_5​

Answer 1

Explanation:

0.02 mole of HCl neutralizes 0.02 moles of NH

. So remaining concentration of NH3

is 0.08 moles per litre.

Now, mixture of 3 4

Cl behaves as buffer.

pOH=pK

b

+log

[NH

3

]

[NH

4

Cl]

=4.74+log

0.08

0.1

=4.83

∴pH=14−4.83=9.16 (final pH)

(ii) 0.02 mole of NaOH neutralizes 0.02 mole of NH

4

Cl. So remaining concentration of NH

4

Cl will be 0.08 moles per litre.

∴pOH=p+ [4 /3]

=4.74+log 0.08/0.1- 4.64

∴ Final pH=14−4.64=9.35

Now initial pH=14−(p+4/[3])

=14−(4.74+0)=9.26

So, case (i) change in pH=9.26−9.16=0.1

case (ii) change in pH=9.35−9.28=0.09