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The pH of 0.08 M HOCI is 2.85. Calculate the degree of dissociation. |
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Answer» Solution :Given value of `pH=2.85, [H^+] = 1.4 XX 10^(-3) mol L^(-1)` `[H^(+)] =C alpha, 1.4 xx 10^(-3)= 0.008 xxalpha,alpha= 1.75xx10^(-2)` The DEGREE of dissociation, `alpha =1.77 xx 10^(-2)`. |
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