Question

The ph of 0.02m solution of ammonia is 10.78 calculate

Answer 1

Answer:

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Explanation:

α=5%=0.05; degree of ionisation / dissociation.

NH

3

+H

2

O⟶NH

4

+

+OH

−

Concentration =0.02M and α=0.05

Therefore, [OH

−

]=0.05×0.02=10

−3

M

pOH=−log[10

−3

]=3

pH=14−pOH=14−3=11