The K_(sp)of PbCO_(3) and MgCO_(3) are 1.5xx10^(-15) and 1xx10^(-15) respectively at 298 K.The concentration of Pb^(2+) ions in a saturated solution containing MgCO_(3) and PbCO_(3) is

The K_(sp)of PbCO_(3) and MgCO_(3) are 1.5xx10^(-15) and 1xx10^(-15) r

1.	The K_(sp)of PbCO_(3) and MgCO_(3) are 1.5xx10^(-15) and 1xx10^(-15) respectively at 298 K.The concentration of Pb^(2+) ions in a saturated solution containing MgCO_(3) and PbCO_(3) is
Answer» `1.5 xx 10^(-8)M` `3xx10^(-8) M` `2xx10^(-8) M` `2.5xx10^(-8)M` Solution :When both `PbCO_(3) and MgCO_(3)` are present in the solution, suppose solubility of `PbCO_(3)` is x mol `L^(-1)` and that of `MgCO_(3)` is y mol `L^(-1)` . Then `{:(PbCO_(3),HARR,Pb^(2),+,CO_(3)^(2-)),(,,x,,x+y):}` `{:(MgCO_(3),hArr,"Mg"^(2),+,CO_(3)^(2-)),(,,x,,x+y):}` `(K_(sp)(PbCO_(3)))/(K_(sp)(MgCO_(3)))=(x(x+y))/(y(x+y))=(x)/(y)` `=(1.5xx10^(-15))/(1.0xx10^(-15))=1.5` Thus, x = 1.5 y Also `K_(sp)(PbCO_(3))=x(x+y)=1.5xx10^(-15)` `:. 1.5 y (1.5 y +y)= 1.5xx10^(-15)` or `3.75 y^(2)=1.5xx10^(-15)` or, `y=((1.5xx10^(-15))/(3.75))^(1//2) = 2 xx 10^(-8)` `x=1.5 y = 1.5 xx (2xx10^(-8))=3xx10^(-8)M`