By adding 20 mL of 0.1 N HCI to 20 mL of 0.001 N KOH, the pH of the obtained solution will be - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
2
B
1.3
C
0
D
7

Solution:

We know that, 20 mL of 0.1 N HCI contains

$[H^{+}] = \frac{20}{1000} \times 0 .1 = 2 \times 10^{- 3} mol Similarly, 20 mL of 0.001 N KOH contains [{OH}^{-}] = \frac{20}{1000} \times 0 .001 = 2 \times 10^{- 5} mol [H^{+}] in the mixture = (2 \times 10^{- 3}) - (2 \times 10^{- 5}) = 1 .98 \times 10^{- 3} [H^{+}] = \frac{1 .98 \times 10^{- 3}}{40 \times 10^{- 3}} = 4.950 \times 10^{- 5}$
$pH = - \log [H^{+}] = - \log (4.950 \times 10^{- 5}) = - \log 4.950 - \log 10^{- 5} = - \log 4.950 - (- 5) \log 10 pH = - 3 .6946 + 5 = 1 .3$

Solution:

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