Calculate the pH of a 7.71×10^{-3} M solution of H_{2}SO_{4}.

(K_{a} = 0.0120 for HSO_{4}^{-})

In a 6.22×10^{-2} M solution, a monoprotic acid, is 27.5% dissociated.

Calculate K_{a} for this acid.

H2SO4 ==== HSO4- + H+

Ka=[HSO4-][H+]/[H2SO4]=0.0120

Assuming the final concentration of H2SO4 is x

Then (7.71×10-3-x)^2/x=0.0120

x^2-7.71×10-3x+7.71^2x10-6=0.0120x

x^2-0.0274x+5.944x10-5=0

x=0.002375

Therefore the final [H+]=7.71×10-3-0.002375=0.005335

pH=-log[H+]=-log(0.005335)=2.27

In a 6.22×10-2 M solution, a monoprotic acid, is 27.5% dissociated.

So Ka=(27.5%x6.22x10-2)^2/6.22×10-2x72.5%=0.275^2x6.22x10-2/0.725=6.49x10^(-3)

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