Question

What will b the result if $100ml\text{of}0.06\text{M}Mg{\left(N{O}_3\right)}_2$ is added to $50ml\text{of 0}\text{.06M}N{a}_2{C}_2{O}_4?$ $\left[K_{sp}\text{of}Mg{C}_2{O}_4=8.6\text{x}{10}^{-5}\right]$

• A. A precipitate will not be formed
• B. A precipitate will form and an excess of $Mg{}^{2+}$ ions will remain in the solution
• C. A precipitate will form and an excess of $C_2{O_4}^{2-}$ ions will remain in the solution
• D. A precipitate is form but neither ion is present in excess

Answer 1

Answer: (B), (C)

The correct options are
B A precipitate will form and an excess of $Mg{}^{2+}$ ions will remain in the solution
C A precipitate will form and an excess of $C_2{O_4}^{2-}$ ions will remain in the solution