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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
Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4}\, M$  $N_2CO_3$ solution. At what concentration of $Ba^{2+}$ will aprecipitate begin to form? ($K_{sp}$  for $BaCO_3 = 5 .1 \times 10^{-9}$)
  • ✓
    $5.1 \times 10 ^{-5}\, M$
  • B
    $8.1  \times 10 ^{-8}\, M$
  • C
    $8.1 \times 10 ^{-7}\, M$
  • D
    $4.1  \times 10 ^{-5}\, M$
✓

Answer

Correct option: A.
$5.1 \times 10 ^{-5}\, M$
a
$N A_{2} C O_{3} \quad \rightarrow \quad 2 N a^{+}\quad +\quad C O_{3}^{2}$

$ {1 \times 10^{-4} M} \quad {1 \times 10^{-4} M} \quad  {1 \times 10^{-4} M} $

$K_{S P\left(B a C O_{3}\right)}=\left[B a^{2+}\right]\left[C O_{3}^{2-}\right]$

$\left[B a^{2+}\right]=\frac{5.1 \times 10^{-9}}{1 \times 10^{-4}}$

$=5.1 \times 10^{-5} M$

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