Post a New Question


posted by on .

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!

  • chemistry - ,

    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.

  • chemistry - ,

    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?

  • chemistry - ,

    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^+.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question