Given : `BaF_(2)`(s) `harr` `Ba^(+2)`(aq)+2`F^(-)`(aq);`K_(sp)`=`1.0times10^(-8)``M^(3)`
HF(aq) `harr` `H^(+)`(aq)+`F^(-)`(aq);`K_(a)`=`6.5times10^(-6)`M The value of `K_(c)` for
`BaF_(2)`(s)+`2H^(+)(aq)` `harr` `Ba^(+2)(aq)`+ 2HF(aq) is:

Equilibrium const. {:(H_2S(aq)hArr H^+(aq)+HS^(-)(aq)," " K_1=9.5xx10^(-8)),(HS^(-)(aq)hArr H^+(aq)+S^(2-)(aq)," " K_2=1.0xx10^(-19)):} What is the equilibrium constant for the reaction ? S^(2-)(aq)+2H^(+)(aq) hArr H_2S(aq) " " K=?

If K_(sp) (PbSO_(4)) =1.8 xx 10^(-8) and K_(a) (HSO_(4)^(Theta)) = 1.0 xx 10^(-2) the equilibrium constant for the reaction. PbSO_(4)(s) +H^(o+)(aq) hArr HSO_(4)^(Theta)(aq) +Pb^(2+)(aq) is

The equilibrium equation and K_(a) values for the three acids are given at 25^(@)C : HA(aq)+H_(2)OhArrH_(3)O^(+)(aq)+A^(-)(aq),K_(a)=2xx10^(-5) HB(aq)+H_(2)OhArrH_(3)O^(+)(aq)+B^(-)(aq),K_(a)=4xx10^(-6) HC(aq)+H_(2)OhArrH_(3)O^(+)(aq)+C^(-)(aq),K_(a)=1xx10^(-4) The pH of 0.2M aqueous HA solution is :

The equilibrium equation and K_(a) values for the three acids are given at 25^(@)C : HA(aq)+H_(2)OhArrH_(3)O^(+)(aq)+A^(-)(aq),K_(a)=2xx10^(-5) HB(aq)+H_(2)OhArrH_(3)O^(+)(aq)+B^(-)(aq),K_(a)=4xx10^(-6) HC(aq)+H_(2)OhArrH_(3)O^(+)(aq)+C^(-)(aq),K_(a)=1xx10^(-4) The pH of 0.1M aqueous NaC solution is : (NaC is sodium salt of acid HC)

AgBr (s) harr Ag^(+) (aq) + Br^(-) (aq); E^(@) = -0.72V If the value of K_(sp) at 25^(@)C is 10^(-b) then the value of "b" is Use (2.303RT)/(F) = 0.06 )