PH of two solutions-I- 50 ml of 0-2 M - HCl -50 ml of 0-2 M HA K a -1-0 - 10-5 andII- 50 ml of 0-2 M HCl -50 ml of 0-2 M NaA will be respectivelyA- 3 and 1B- 0-70 and 2-85C- 1 and 2-85D- 1 and 3

The correct option is D 1 and 3 (I) [HCl] in the mixture = 50 × 0.2/100 = 0.1 M In presence of strong acid HCl, the weak acid HA remains partically unionized. ...

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Standard XII

Chemistry

Ostwald Dilution Law Mathematical Interpretation

pH of two sol...

Question

pH of two solutions :
I. 50 ml of 0.2 M~HCl + 50 ml of 0.2 M HA $(K_{a} = 1.0 \times 10^{- 5})$ and
II. 50 ml of 0.2 M HCl + 50 ml of 0.2 M NaA will be respectively

0.70 and 2.85

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1 and 2.85

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1 and 3

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3 and 1

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Solution

The correct option is C
1 and 3

(I) [HCl] in the mixture $= \frac{50 \times 0.2}{100} = 0.1 M$
In presence of strong acid HCl, the weak acid HA remains partically unionized. Hence
$[H^{+}] = [H^{+}]_{H C l} = 0.1 M, p H = 1$
(II) $N a A + H C l \to N a C l + H A;$
$[H A] = \frac{50 \times 0.2}{100} = 0.1 M$
No HCl is left. Hence $[H^{+}]$
$= \sqrt{K_{a} C} = \sqrt{1.0 \times 10^{- 5} \times 0.1} = 1.0 \times 10^{- 3}; p H = 3$

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pH of two solutions : I. 50 ml of 0.2 M~HCl + 50 ml of 0.2 M HA (Ka=1.0×10−5) and II. 50 ml of 0.2 M HCl + 50 ml of 0.2 M NaA will be respectively

The correct option is C 1 and 3 (I) [HCl] in the mixture =50×0.2100=0.1M In presence of strong acid HCl, the weak acid HA remains partically unionized. Hence [H+]=[H+]HCl=0.1M,pH=1 (II) NaA+HCl→NaCl+HA; [HA]=50×0.2100=0.1M No HCl is left. Hence [H+] =√KaC=√1.0×10−5×0.1=1.0×10−3;pH=3

pH of two solutions :
I. 50 ml of 0.2 M~HCl + 50 ml of 0.2 M HA $(K_{a} = 1.0 \times 10^{- 5})$ and
II. 50 ml of 0.2 M HCl + 50 ml of 0.2 M NaA will be respectively