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 What is the molar solubility of Cu(OH)_(2) , in 1.0 M NH_(3) if the deep blue complex ion [Cu(NH_(3))_(4)]^(2+) is formed. The K_(sp), of Cu(OH)_(2) , is 1.6xx 10^(-19) and K_(3) , of [Cu(NH_(3))_(4) is 1.1 xx 10^(13)
The simultaneous solubility of Ag CN (K_(f) =2.5 xx 10^(-16) ) and AgCl (Ksp =1.6 xx 10^(-10 ) ) in 1. 0M NH_3 (aq) are respectively : [Given ,K_f [Ag(NH_3)_2^(+) ]= 10^(7) ]
Sparingly soluble salts maintains their solubility product value in their saturated solutions irrespective of the sources of the ions . what is the molar solubility of AgNO_3 in a 0.1 M H_2S solution buffered at pH= 2 (K_1 and K_2 " of " H_2S " are " 10^(4) and 10^(-8) respectively) (K_(sp) " of " Ag_2S = 4xx 10^(-13) ) (Note : No Ag_2S precipitate should be formed)
If the solubility product of Ni(OH)_2 is 4.0 xx 10^(-15) the solubility (in mol L^(-1) ) is
Calculate the solubility of A_(2)X_(2) in pure water, assuming that neither kind of ion reacts with water. The solubility product of A_(2)X_(3) , K_(sp)=1.1xx10^(-23) .
Calculate the molar solubility of Ni(OH)_(2) in 0.10M NaOH . The ionic product of Ni(OH)_(2) is 2.0xx10^(-15)
The solubility of AgCl in 0.1M NaCI is (K_(sp) " of AgCl" = 1.2 xx 10^(-10))
NARENDRA AWASTHI-IONIC EEQUILIBRIUM-Assertin-Reason Type Questions
