Verified Solution • 2025–26 Session moderate

QJEEM07Ch20f6: JEE Main 2007 Chemistry - Class 11 JEEM07Ch20f6 Solution: In a sautrated solution of the sparingly soluble strong…

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) + IO3−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?

Multiple Choice Options

  • A 28.3 × 10−2 g
  • B 2.83 × 10−3 g
  • C 1.0 × 10−7 g
  • D 1.0 × 10−4 g

Step-by-Step Verified Solution

Let    s=   \,\,\,s = \,\,\, solubility

AgIO3 ⇌ Ag+s  IO3−sAg{\rm I}{O_3}\,\rightleftharpoons\,\mathop {A{g^ + }}\limits_s \,\,\mathop {{\rm I}{O_3}^ - }\limits_s

Ksp=[Ag+][IO3−]{K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{\rm I}{O_3}^ - } \right]

=s×s=s2= s \times s = {s^2}

Given    Ksp=1×10−8\,\,\,{K_{sp}} = 1 \times {10^{ - 8}}

∴\therefore    s=Ksp=1×10−8\,\,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}}

=1.0×104  mol/lit= 1.0 \times {10^4}\,\,mol/lit

=1.0×10−4×283 g/lit= 1.0 \times {10^{ - 4}} \times 283\,g/lit

(as Molecular mass of Ag IO3=283Ag\,{\rm I}{O_3} = 283 )

=1.0×10−4×283×1001000gm/100ml= {{1.0 \times {{10}^{ - 4}} \times 283 \times 100} \over {1000}}gm/100ml

=2.83×10−3 gm/100 ml= 2.83 \times {10^{ - 3}}\,gm/100\,ml

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