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6-2.Equilibrium-II (Ionic Equilibrium)
medium
ઘન $Ba(NO_3)_2$ $1.0 \times 10^{-4}\, M\, Na_2CO_3$ દ્રાવણમાં ધીમે ધીમે ઓગળે છે. $Ba^{2+}$ની કઈ સાંદ્રતા એ $BaCO_3$ના અવક્ષેપ શરૂઆતથી બનશે?($BaCO_3 = 5.1 \times 10^{-9}$ માટે $K_{sp}$)
A
$5.1 \times {10^{ - 5}}\,M$
B
$7.1 \times {10^{ - 8}}\,M$
C
$4.1 \times {10^{ - 5}}\,M$
D
$8.1 \times {10^{ - 7}}\,M$
(JEE MAIN-2013)
Solution
Given $N{a_2}C{O_3} = 1.0 \times {10^{ – 4}}\,M$
$\therefore \,[CO_3^ – ] = 1.0 \times {10^{ – 4}}\,M$
$i.e.\,\,\,s = 1.0 \times {10^{ – 4}}\,M$
At equilibrium
$[B{a^{ + + }}][CO_3^ – ] = {K_{sp}}\,of\,BaC{O_3}$
$[B{a^{ + + }}] = \frac{{{K_{sp}}}}{{[CO_3^ – ]}} = \frac{{5.1 \times {{10}^{ – 9}}}}{{1.0 \times {{10}^{ – 4}}}}$
$ = 5.1 \times {10^{ – 5}}\,M$
Standard 11
Chemistry
