What will be the result if 100mL of 0.06M $M g {(N O_{3})}_{2}$ is added to 50mL of 0.06M $N a_{2} C_{2} O_{4}$ ? $[K_{s p} o f M g C_{2} O_{4} = 8.6 \times 10^{- 5}]$

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A precipitate will not be formed

A precipitate will form and an excess of $M g^{+ 2}$ ions will remain in the solution

A precipitate will form and an excess of $C_{2} O_{4}^{2 -}$ ions will remain in the solution

A precipitate will form but neither ion is present in excess

answer is B.

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Detailed Solution

$[M g^{+ 2}] = 0.06 M \Rightarrow [C_{2} O_{4}^{- 2}] = 0.06 M A f t e r m i x i n g M_{1} V_{1} = M_{2} V_{2} [M g^{+ 2}] = 0.04 M [C_{2} O_{4}^{- 2}] = 0.02 M M g^{+ 2} + C_{2} O_{4}^{- 2} \overset{}{⇌} M g C_{2} O_{4} \begin{array}{l} 0.04 0.02 0 \\ 0.02 - 0.02 \end{array} ∴ K_{s p} < I P p p t w i l l f o r m W i t h e x c e s s M g^{+ 2} i o n s i n s o l u t i o n$