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Stoichiometric Calculations

10-2 mole of ...

Question

$10^{- 2}$ mole of NaOH was added to 10 litre of water the pH will change by

A

4

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B

3

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C

11

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D

7

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Solution

The correct option is A 4
old pH = 11
New [OH] = $10^{- 2} \times \frac{1}{10} = 10^{- 3}$
New pH = 11
Change in pH = 4

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