Respuesta :
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
[tex][Ca^{2+}] = 2.1 \times 10^{-5}[/tex] M
[tex][C_{2}O_{4}^{2-}] = 4.75 \times 10^{-5}[/tex] M
[tex]K_{sp} (CaC_{2}O_{4}) = 2.3 \times 10^{-9}[/tex]
As expression for [tex]K_{sp}[/tex] will be as follows.
[tex]K_{sp} = [Ca^{2+}][C_{2}O^{2-}_{4}][/tex]
= [tex]2.3 \times 10^{-9}[/tex]
And, expression for ionic product will be as follows.
Q = [tex][Ca^{2+}][C_{2}O^{2-}_{4}][/tex]
= [tex]2.1 \times 10^{-5} \times 4.75 \times 10^{-5}[/tex]
= [tex]9.975 \times 10^{-10}[/tex]
Since, it is shown here that ionic product is less that solubility product. Hence, no precipitate will form.
Thus, we can conclude that nothing will happen since Ksp > Q for all possible precipitants.
When these solutions are mixed then,
B. Nothing will happen since Ksp > Q for all possible precipitants.
Given:
Ksp ([tex]CaC_2O_4[/tex]) = [tex]2.3 *10^{-9}[/tex]
Calcium ion = [tex]2.1 * 10^{-5} M[/tex]
Oxalate ion = [tex]4.75 * 10^{-5} M[/tex]
Solubility product:
[tex]K_{sp}=[Ca^{2+}][C_2O_4^{2-}]=2.3*10^{-9}[/tex]
And, expression for ionic product will be as follows.
[tex]Q=[Ca^{2+}][C_2O_4^{2-}]\\\\Q=2.1*10^{-5}*4.75*10^{-5}\\\\Q=9.975*10^{-10}[/tex]
Thus, ionic product is less that solubility product. Hence, no precipitate will form.
Thus, we can conclude that nothing will happen since Ksp > Q for all possible precipitants. So correct option is B.
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