1.

What will be the value of pH of 0.01 mol `dm^(-3) cH_(3)CO OH (K_(a) = 1.74 xx 10^(-5))`?A. 3.4B. 3.6C. 3.9D. `3.0`

Answer» Correct Answer - A
`pH = (1)/(2) (pK_(a)-log c)=(1)/(2) [-log (1.74xx10^(-5))-log 10^(-2)]`
`=(1)/(2) [ (5-0.2405) + 2] = (1)/(2) (6.76) = 3.38 ~= 3.4`.


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