Question

It has been found that the pH of a 0.02 M solution of an acid is 4.15. calculate the concentration of the anion , ionisation constant of the acid and its pKa​

Answer 1

Answer:

pH=−log[H+]=4.15

[H+]=antilog(−4.15)=7.08×10−5

[A−]=[H+]=7.08×10−5

The concentration of undissociated acid is 0.01−0.000071=0.009929M.

HA+H2O⇌H3O++A−

Ka=[HA][H3O+][A−]=0.009929(7.08×10−5)(7.08×10−5)=5.05×10−7

pKa=−logKa=−log5.05×10−7≈6.3

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