First slide

PH of strong acids and strong bases

Question

By adding 20 ml 0.1 N HCl to 20 ml 0.001 N KOH, the pH of the obtained solution will be

Moderate

A

2
B

1.3
C

0
D

7

Solution

20 ml of 0.1 NHCl = $\frac{0.1}{1000} \times 20 g eq = 2 \times 10^{- 3} g eq .$
20 ml of 0.001 KOH = $\frac{0.001}{1000} \times 20 g e q$
$= 2 \times 10^{- 5} g e q .$
$∴ HCl left unneutralised = 2 (10^{- 3} - 10^{- 5}) = 2 \times 10^{- 3} (1 - 0.01) = 2 \times 0.99 \times 10^{- 3} = 1.98 \times 10^{- 3} g eq .$
Volume of solution = 40 ml
$∴ [HCl] = \frac{1.98 \times 10^{- 3}}{40} \times 1000 M = 4.95 \times 10^{- 2} ∴ pH = 2 - \log 4.95 = 2 - 0.7 = 1.3$

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