MCQ
Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-3} \,M \,Na_2CO_3$ solution. At what concentration of $Ba^{2+}$ will be precipitate begin to form ($K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)
  • A
    $4.1 \times 10^{-5} \,M$
  • $5.1 \times 10^{-6} \,M$
  • C
    $8.1 \times 10^{-8} \,M$
  • D
    $8.1 \times 10^{-7} \,M$

Answer

Correct option: B.
$5.1 \times 10^{-6} \,M$
b
$5.1 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right]\left[10^{-3}\right] \Rightarrow\left[\mathrm{Ba}^{+2}\right]$

$=5.1 \times 10^{-6} \,\mathrm{m}$

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