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20mL of 0.1M(H(2)SO(4))solution is added...

20mL of `0.1M(H_(2)SO_(4))`solution is added to 30mL of `0.2M(NH_(4)OH)` soulution. The `._(p)H` of the resultant mixture is : `[._(p)k_(b) " of " NH_(4)OH=4.7]`.

A

9.4

B

9

C

5

D

5.2

Text Solution

Verified by Experts

The correct Answer is:
B
`H_(2)SO_(4) + 2NH_($)OH to (NH_(4))_(2)SO_(4) + 2H_(2)O`
Initial 2m mol 6m mol –
Initial – 2m mol 2m mol
Total volume 20ml + 30ml = 50ml
`[NH_(4)OH] =2/50 M`
`[NH_(4)^(+)=2xx 2/50 = 4/50 M`
`pOH =pK_(b) + "log"(([NH_(4)^(+)])/([NH_(4)OH)])=4.7 + "log" (4/50)/(2//50)=4.7 + log2 approx 5`
`therefore pH=14-5=9`
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