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100mL of " 0.1 M NaOH" solution is titra...

`100mL` of `" 0.1 M NaOH"` solution is titrated with `100mL` of `"0.05 M "H_(2)SO_(4)` solution. The `pH` of the resulting solution is `: (` For `H_(2)SO_(4), K_(a1)=10^(-2))`

A

5.36

B

8.64

C

9.26

D

4.74

Text Solution

Verified by Experts

The correct Answer is:
B
`2NH_(4)OH+H_(2)SO_(4)to((NH_(4))_(2)SO_(4)+2H_(2)O`
Initial `1m` mol, `0.4m` mol, 0
`0.2m "mol" 0, 0.4 m "mol'`
`[NH_(4)^(+)]=(2xx0.4)/14, [NH_(4)OH]=0.2/14`
`:.pOH=pK_(b)+log(0.8//14)/(0.2//14)`
`=4.76+log4=4.76+0.6`
`=5.36`
`pH=14-5.36=8.64`
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