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a
8.0
b
6.0
c
6.98
d
7.02
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Correct option is D
[NaOH]=10−8M OH−=10−8M H+=10−6M pH ≠6 since, solution is basic OH− of water in this dilute solution should be added ∴ OH−total =1×10−7M⏟H2O+1×10−8M⏟NaOH=11×10−8 ∴ pOH=8−log11 pH=14−pOH =6+log11=7.02Similar Questions
Milliequivalent of Ba(OH)2 present in 100 mL solution to have pH = 13, is
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