First slide

Concentration methods

Question

10 ml of 0.1 M NaOH solution is diluted to 1000 ml solution. Proton concentration of the solution at 298 K will be

Easy

A

$10^{- 11} M$
B

$5 \times 10^{- 10} M$
C

$5 \times 10^{- 12} M$
D

$2 \times 10^{- 11} M$

Solution

Dilution law :
$V_{1} N_{1} = V_{2} N_{2} 10 (0.1) 2 = 1000 \times N_{2} N_{2} = 2 \times 10^{- 3} N; [{OH}^{-}] = Normality of strog base ∴ [{OH}^{-}] = 2 \times 10^{- 3} M; Thus [H^{+}] = \frac{10^{- 14}}{2 \times 10^{- 3}} = 5 \times 10^{- 12} M;$

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