1.

100 ml 0.2 M `Na_(2)A` is titrated with 100ml 0.03 M HCl. Calculate pH of final solution. `pKa_(1)(H_(2)A)=5" "pKa_(2)(H_(2)A)=9`A. 2B. 7.5C. 5D. 9

Answer» Correct Answer - A
`SO_(2)Cl_(2)(g)rarrSO_(2)(g)+Cl_(2)(g)`
0.5
`0.5-x " "x" "x`
`K_(A)=(1)/(100)ln.(0.5)/(0.25)=(ln2)/(100)`
Rate `=KP_(A) = (ln2)/(100)xx0.3=2.1xx10^(-3)"atm"//"sec"`


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