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`10^(-2)` mole of `NaOH` was added to `10 litres` of water. The `pH` will change byA. 4B. 7C. 5D. 8 |
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Answer» Correct Answer - A `(OH)(0.01)/(10)=10^(-3)rArrpOHrArrpH=14-3=11` change in pH = 11-7=4 |
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