What is the pH of a solution in which 25 mL of 0.1 MNaOH is added to 25 mL of 0.08 MHCl and final solution is diluted to 500 mL?
A
3
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B
11
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C
12
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D
13
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Solution
The correct option is B11 Milli-equivalents of HCl=25×0.08=2
Milli-equivalents of NaOH=25×0.1=2.5
As milli-equivalents of NaOH is more than milli-equivalents of HCl, then the remaining concentration of [OH−]=2.5−2500=10−3 [H+]=10−11 pH=−log(10−11)=11