What will be the result if 100 mlL of 0.06 M Mg`(NO_(2))_(2)` is added to 50 ml. of 0.06 M ` Na_(2) C_(2) O_(4)` ? [ Ksp of `MgC_(2) O_(4) = 8.6 xx 10^(-5) ` ]

Options

A. A precipitate will not be formed
B. A precipitate will form and an excess of `Mg^(2+) ` ions will remains in the solution
C. A precipitate will form and an excess of `C_2O_4^(2-) ` ions will remain in the solution
D. A precipitate will form but neither ion is present in excess

Answer

B

Explanation

`[Mg^(+2) ] =0.06 M rArr [C_2O_4^(-2) ] =0. 06 M`
But after mixing
` M_1V_1 = M_2V_2 rArr 0. 0 6 xx 100 = M_2 xx 150`
` [Mg^(+2)] =0.04 M rArr M_1 V_1 =M_2 V_2`
` 0.06 xx 50 = M_2 xx 150 rArr [C_2 O_4^(-2) ]= 0.02 M`
` {:( Mg^(+2) +, C_2O_4^(-2) hArr, MgC_2O_4),( 0.04 M , 0.02M , 0) ,( 0.02 , -,0.02M), (0.02M , , ):}`
` I.P =[Mg^(+2) ] [C_2O _4^(-2)]`
` =(4xx 10 ^(-2) ) (2xx 10 ^(-2) ) = 8 xx 10 ^(_4) `
` therefore K_(sp) lt I. P ` , ppt will from