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6.2. Equilibrium - II (icon Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.2. Equilibrium - II (icon Equilibrium)1 Mark
MCQ
$K_{SP}$ of $AgCl$ at $18\,^oC$ is $1.8 \times 10^{-10}$. If concentration of $Ag^+$ is $4 \times 10^{-3}\, mol/litre$ the concentration of $Cl^-$ that is required for $AgCl$ precipitation
  • ✓
    $4.5 \times 10^{-8}\, mol/l$
  • B
    $4 \times 10^{-3}\, mol/l$
  • C
    $7.2 \times 10^{-13}\, mol/l$
  • D
    $4.5 \times 10^{-7}\, mol/l$
✓

Answer

Correct option: A.
$4.5 \times 10^{-8}\, mol/l$
a
$AgCl \rightleftharpoons Ag ^{+}+ Cl ^{-}$

$\quad\quad\quad 4 \times 10^{-3} \quad x$

$K_{s p}=\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]$

$\Rightarrow 1.8 \times 10^{-10}=4 \times 10^{-3} \times x$

$\Rightarrow \quad x =4.5 \times 10^{-8}$

$\therefore {\left[ Cl ^{-}\right]=4.5 \times 10^{-8}\, m . }$

So, if the concentration of $Cl ^{-}$ is greater than $4.5 \times 10^{-8} \,m$, ionic product $(Q)$ will increase, And if $Q$ is greater than $K_{s p}, Cl^{-}$ will precipitate.

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