1.

`10^(-2)` mole of `NaOH` was added to `10 litres` of water. The `pH` will change byA. 4B. 7C. 5D. 8

Answer» Correct Answer - A
`(OH)(0.01)/(10)=10^(-3)rArrpOHrArrpH=14-3=11`
change in pH = 11-7=4


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