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Question

The pH of solution by mixing 50 mL of 0.6 M NH4OH and 50 mL of 0.1M H2SO4 is : Given Ka for NH+4=109


A
4.39
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B
6.61
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C
9.3
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D
7.39
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Solution

The correct option is C 9.3
2NH4OH+H2SO4(NH4)2SO4+2H2O
Meg.at t=0301000Meq.after reaction2001010
  [(NH4)2SO4]=12×100M
  [NH+4]=10×22×100M=0.1M
[NH4OH]=20100=0.2M
pOH=PKb+log[NH+4][NH4OH]=105=log0.10.2=50.3010=4.6989
 pH=9.3010

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