K_(sp) of M(OH)_(x) , is 27 xx 10^(-12) and its solubility in water is 10^(-3) mol litre^(-1) . Find the value of X
The solubility product of BaSO_(4) at 18^(0) C is 1.5xx10^(-9) . It solubility (mole lit^(-1)) at the same temperature is
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) ,
At 298 K , the K_(sp) value of Fe(OH)_3 in aqueous solution is 3.8 xx 10^(-38) . The solubility of Fe^(3+) ions will increase when
At 298 K , the K_(sp) of M_(2) SO_(4) is 3.2 xx 10^(-5) . The maximum concentration of SO_(4)^(2-) ion that could be attained in a saturated solution of this solid at 298 K is
K_(sp) (BaSO_(4)) is 1.1 xx 10^(10). In which case is BaSO_(4) precipitated ?
The solubility of CaF_(2) is 2xx10^(-4) mole /litre . Its solubility product is
NARENDRA AWASTHI-IONIC EEQUILIBRIUM-Assertin-Reason Type Questions
