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6-2.Equilibrium-II (Ionic Equilibrium)
medium
The solubility of $BaS{O_4}$ in water is $2.33 \times {10^{ - 3}}\,gm/litre.$ Its solubility product will be (molecular weight of $BaS{O_4} = 233)$
A
$1 \times {10^{ - 5}}$
B
$1 \times {10^{ - 10}}$
C
$1 \times {10^{ - 15}}$
D
$1 \times {10^{ - 20}}$
(AIIMS-1998)
Solution
(b) The solubility of $BaS{O_4}$ in g/litre is given $2.33 \times {10^{ – 3}}$
in mole/litre.$n = \frac{W}{{m.\,wt}}$$ = 1 \times {10^{ – 5}}$$ = \,\frac{{2.33 \times {{10}^{ – 3}}}}{{233}}$
Because $BaS{O_4}$ is a compound
${K_{sp}} = {S^2} = {[1 \times {10^{ – 5}}]^2}$ $ = 1 \times {10^{ – 10}}$
Standard 11
Chemistry