JEE MAIN - Chemistry (2009 - No. 6)
Solid Ba(NO3)2 is gradually dissolved in a 1.0 $$\times$$ 10-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form ?
(Ksp for BaCO3 = 5.1 $$\times$$ 10−9 )
(Ksp for BaCO3 = 5.1 $$\times$$ 10−9 )
5.1 $$\times$$ 10-5 M
8.1 $$\times$$ 10-8 M
8.1 $$\times$$ 10-7 M
4.1 $$\times$$ 10-5 M
Explanation
$$\mathop {N{a_2}C{O_3}}\limits_{1 \times {{10}^{ - 4}}M} \to \mathop {2N{a^ + }}\limits_{1 \times {{10}^{ - 4}}M} \,\, + \,\,\mathop {C{O_3}^{2 - }}\limits_{1 \times {{10}^{ - 4}}M} $$
$${K_{SP\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right]$$
$$\left[ {B{a^{2 + }}} \right] = {{5.1 \times {{10}^{ - 9}}} \over {1 \times {{10}^{ - 4}}}}$$
$$ = 5.1 \times {10^{ - 5}}M$$
$${K_{SP\left( {BaC{O_3}} \right)}} = \left[ {B{a^{2 + }}} \right]\left[ {CO_3^{2 - }} \right]$$
$$\left[ {B{a^{2 + }}} \right] = {{5.1 \times {{10}^{ - 9}}} \over {1 \times {{10}^{ - 4}}}}$$
$$ = 5.1 \times {10^{ - 5}}M$$
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