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6-2.Equilibrium-II (Ionic Equilibrium)
medium
एक अम्ल $HA $ का वियोजन स्थिरांक $1 \times {10^{ - 5}}$ है । अम्ल के $0.1 $ मोलर विलयन की $pH$ होगी
A
$5$
B
$4$
C
$3$
D
$1$
Solution
(c) $HA \rightleftharpoons {H^ + } + {A^ – }$
$[{H^ + }] = 0.1\,\,M$; ${[{H^ + }]^2} = {K_a} \times C$
$[{H^ + }] = \sqrt {{K_a} \times C} = \sqrt {1 \times {{10}^{ – 5}} \times 0.1} = \sqrt {{{10}^{ – 6}}} $
$[{H^ + }] = {10^{ – 3}}M;\,\,pH = 3$
Standard 11
Chemistry
