Chemistry, asked by jsubashini5162, 10 months ago

A monoprotic acid in a 0.1 m solution ionises to 0.001

Answers

Answered by Anonymous
7

Explanation:

HA = H+ + A-

% Ionization = [H+] / HA *100

0.001 = [H+] / 0.1*100

[H+]= 1.0^10^-6

ionisation constant = [H+] [A-]/[HA]

= 1.0^10^-6 *1.0^10^-6 /0.1

= 1.0*10^-11

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