This requires a little imagination.
If I call k1 = Ka for HCN it will be
HCN ==> H^+ + CN^-
k1 = Ka for HCN = (H^+)(CN^-)/(HCN)
k2=Ka2 for H2SO4 HSO4^- ==> H^+ + SO4^-
k2 = (H+)(SO4^2-)/(HSO4^-) but I want the reverse of this; therefore,
1/k2 = (HSO4^-)/(H^+)(SO4^2-)
HCN + SO4^- ==> HSO4^2- + CN^-
k1*(1/k2) = [(H^+)(CN^-)/(HCN)]*[(HSO4^-)/(H^+)(SO4^2-)] and this leaves
(CN^-)((HSO4^-)/(HCN)(SO4^2-) which is Keq for the reaction you started with.
So k1*(1/k2) = look in you book for k1 and k2. My text lists (but you want the use the ones listed in your text) = 1E-9/0.012 = about 8E-8 which tells you that the reactants are favored. That is, the equilibrium lies (far) to the left.
For each of the following reactions, predict whether the equilibrium lies predominantly to the left or to the right.
HCN(aq) + SO42-(aq) CN -(aq) + HSO4-(aq)
I think it has something to do with the Ka but im not sure!
3 answers
thanks! also for this question Calculate the hydronium ion concentration and pH in a 0.037 M solution of sodium formate, NaHCO2.
hydronium ion concentration I made an equation: k=x^2/.037-x but when i look up the k what compound am i looking for? how do i know?
hydronium ion concentration I made an equation: k=x^2/.037-x but when i look up the k what compound am i looking for? how do i know?
If you wrote the hydrolysis equation (the one form Kb formate) you would know.
........HCO2^- + HOH ==> HCOOH + OH^-
Kb(for formate) = (Kw/Ka for HCOOH) = (x)(x)/(HCO2^-) and solve for x = OH^- then convert to pH. Don't forget this is OH you're solving for, not H^+.
........HCO2^- + HOH ==> HCOOH + OH^-
Kb(for formate) = (Kw/Ka for HCOOH) = (x)(x)/(HCO2^-) and solve for x = OH^- then convert to pH. Don't forget this is OH you're solving for, not H^+.