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JEE Main 2003
Exam
JEE Main
Year
2003
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2003 · Chemistry · Ionic Equilibrium

The solubility in water of a sparingly soluble salt AB is 1.0 × 10^...

The solubility in water of a sparingly soluble salt AB\mathrm{AB} is 1.0×10−5 mol L−11.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}. Its solubility product number will be

Options

  1. A4×10−104 \times 10^{-10}
  2. B1×10−151 \times 10^{-15}
  3. C1×10−101 \times 10^{-10}
  4. D4×10−154 \times 10^{-15}

Correct answer

D. 4×10−154 \times 10^{-15}

Step-by-step solution

[A]=1.0×10−5,[ B]=[1.0×10−5]Ksp=[2.B]2[ A]=[2×10−5]2[1.0×10−5]=4×10−15\begin{array}{ll}{[\mathrm{A}]=1.0 \times 10^{-5},} & {[\mathrm{~B}]=\left[1.0 \times 10^{-5}\right]} \\ \mathrm{K}_{\mathrm{sp}}=[2 . \mathrm{B}]^2[\mathrm{~A}] \quad=\left[2 \times 10^{-5}\right]^2\left[1.0 \times 10^{-5}\right]=4 \times 10^{-15}\end{array}
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