For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-)] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals Potussium chromate is slowly aded toa solution containing 0.20M Ag NO_(3), and 0.20M Ba(NO_(3))_(2). Describe what happensif the K_(sp) for Ag_(2),CrO_(4), is 1.1 xx 10^(-12) and the K_(sp) of BaCiO_(4), is 1.2 xx 10^(-10),

For a general reaction given below, the value of solubility product ca

1.	For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-)] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals Potussium chromate is slowly aded toa solution containing 0.20M Ag NO_(3), and 0.20M Ba(NO_(3))_(2). Describe what happensif the K_(sp) for Ag_(2),CrO_(4), is 1.1 xx 10^(-12) and the K_(sp) of BaCiO_(4), is 1.2 xx 10^(-10),
Answer» The ` Ag_2CrO_4` pecipitates first out of solution and then `BaCrO_4 ` preciptates. The `BaCrO_4` pecipitates first out of solution and then ` Ag_2CrO_4`preciptates Both `Ag_2CrO_4 and BaCrO_4`PRECIPITATE simultaneously out of solution NEITHER `Ag_2CrO_4 " nor " BaCrO_4` precipitates Solution :` [CrO_4^(2-)]_(Ag^(+))=(Ksp)/( [Ag^(+) ]^(2)) =(1.1xx 10^(-12))/(0.2)=5.5 xx 10 ^(-12) ` ` [CrO_4^(2-) ]_(BA^(+2) ) =(Ksp)/([Ba^(+2)])=(1.2xx 10^(-10))/( 0.2 0) = 6 xx 10^(-10) ` ` therefore Ag_2 Cr O_4 `ppts first