The pH of the solution containing 10mL of 0.1NNaOH and 10mL of 0.05NH2SO4 would be:
A
\N
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B
1
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C
>7
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D
7
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Solution
The correct option is C >7 milliequivalent of NaOH=10×0.1=1 milliequivalent ofH2SO4=10×0.05=0.5 Left milliequivalent of NaOH=1−0.5=0.5M ∴[OH−]=0.510+10=2.5×10−2
we know the relation pOH=−log[OH−]=−log(2.5×10−2)=1.6020
Also, pH=14−pOH pH=14−1.6020=12.398