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Calculate the degree og ionisation of `0.04 M HOCl` solution having ionisation constant `1.25 xx 10^(-4)` ?A. `0.025`B. `0.5`C. `0.25`D. `0.055` |
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Answer» Correct Answer - D `{:(HOCl,hArr,OH^(+),+,Cl^(+),),(C,,,,,),(C (a - alpha),,C alpha,,C alpha,):}` `K_(a) = (c alpha^(2))/(1 - alpha) = (0.04 xx alpha^(2))/(1 - alpha) " " [{:("As",,alpha lt lt lt 1),(because,,1-alpha=1):}]` `1.25 xx 10^(-4) = 0.04 alpha^(2)` `alpha^(2) = (1.25 xx 10^(-4))/(0.04) = (125 xx 10^(-4))/(4)` `alpha = (11)/(2) xx 10^(-2) = (0.11)/(2) = 0.055` |
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