MCQ
Solid $Ba(NO_3)_2$ is gradually dissolved in $1.0 \times 10^{-4} \,M \,Na_2CO_3$  solution. At which concentration of $Ba^{2+}$ the precipition will begin? ( $K_{sp}$ $BaCO_3 = 5.1 \times 10^{-9}$)
  • A
    $4.1 \times 10^{-5} \,M$
  • $5.1 \times 10^{-5} \,M$
  • C
    $8.1 \times 10^{-8} \,M$
  • D
    $8.1 \times 10^{-7} \,M$

Answer

Correct option: B.
$5.1 \times 10^{-5} \,M$
b
$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{CO}_{3}^{-}\right]$

$5.1 \times 10^{-9}=\left[\mathrm{Ba}^{2+}\right] \times 1 \times 10^{-4}$

$5.1 \times 10^{-5}=\left[\mathrm{Ba}^{2+}\right]$

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