JEE MAIN - Chemistry (2007 - No. 4)
In a sautrated solution of the sparingly soluble strong electrolyte AgIO3 (Molecular mass = 283) the
equilibrium which sets in is
AgIO3(s) $$\leftrightharpoons$$ Ag+(aq) + $$IO_3^-$$
If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 $$\times$$10−8, what is the mass of AgIO3 contained in 100 ml of its saturated solution?
AgIO3(s) $$\leftrightharpoons$$ Ag+(aq) + $$IO_3^-$$
If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 $$\times$$10−8, what is the mass of AgIO3 contained in 100 ml of its saturated solution?
28.3 × 10−2 g
2.83 × 10−3 g
1.0 × 10−7 g
1.0 × 10−4 g
Explanation
Let $$\,\,\,s = \,\,\,$$ solubility
$$Ag{\rm I}{O_3}\,\rightleftharpoons\,\mathop {A{g^ + }}\limits_s \,\,\mathop {{\rm I}{O_3}^ - }\limits_s $$
$${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{\rm I}{O_3}^ - } \right]$$
$$ = s \times s = {s^2}$$
Given $$\,\,\,{K_{sp}} = 1 \times {10^{ - 8}}$$
$$\therefore$$ $$\,\,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}} $$
$$ = 1.0 \times {10^4}\,\,mol/lit$$
$$ = 1.0 \times {10^{ - 4}} \times 283\,g/lit$$
(as Molecular mass of $$Ag\,{\rm I}{O_3} = 283$$ )
$$ = {{1.0 \times {{10}^{ - 4}} \times 283 \times 100} \over {1000}}gm/100ml$$
$$ = 2.83 \times {10^{ - 3}}\,gm/100\,ml$$
$$Ag{\rm I}{O_3}\,\rightleftharpoons\,\mathop {A{g^ + }}\limits_s \,\,\mathop {{\rm I}{O_3}^ - }\limits_s $$
$${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{\rm I}{O_3}^ - } \right]$$
$$ = s \times s = {s^2}$$
Given $$\,\,\,{K_{sp}} = 1 \times {10^{ - 8}}$$
$$\therefore$$ $$\,\,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}} $$
$$ = 1.0 \times {10^4}\,\,mol/lit$$
$$ = 1.0 \times {10^{ - 4}} \times 283\,g/lit$$
(as Molecular mass of $$Ag\,{\rm I}{O_3} = 283$$ )
$$ = {{1.0 \times {{10}^{ - 4}} \times 283 \times 100} \over {1000}}gm/100ml$$
$$ = 2.83 \times {10^{ - 3}}\,gm/100\,ml$$
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