Description
Calculate the pH of a 7.71×10-3 M solution of H2SO4.
(Ka = 0.0120 for HSO4-)
In a 6.22×10-2 M solution, a monoprotic acid, is 27.5% dissociated.
Calculate Ka for this acid.
Explanation & Answer
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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