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100 ml 0.2 M Na(2)A is titrated with 10...

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

2

B

7.5

C

5

D

9

Text Solution

Verified by Experts

The correct Answer is:
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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