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10^(-2) mole of NaOH was added to 10 lit...

`10^(-2)` mole of `NaOH` was added to `10 litres` of water. The `pH` will change by

A

4

B

7

C

5

D

8

Text Solution

Verified by Experts

The correct Answer is:
A
`(OH)(0.01)/(10)=10^(-3)rArrpOHrArrpH=14-3=11`
change in pH = 11-7=4
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