Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 × 10^{-4}, M N₂CO₃ solution. At what

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6-2.Equilibrium-II (Ionic Equilibrium)

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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 $BaCO_3 = 5 .1 \times 10^{-9}$)

A

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

B

$8.1 \times 10 ^{-8}\, M$

C

$8.1 \times 10 ^{-7}\, M$

D

$4.1 \times 10 ^{-5}\, M$

(AIEEE-2009)

Solution

$N A_{2} C O_{3} \quad \rightarrow \quad 2 N a^{+}\quad +\quad C O_{3}^{2}$

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

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

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

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

Standard 11

Chemistry

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