Chemistry

What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10^-10?

I am not sure how to this question... I did 6.2x10^-10/ 1.24?? I am not sure if it is right OR are we suppose to find the Kb value which i already did Kb= kw/ka = 1.0x10^-14/6.2x10^-10 = 1.61x10^-5 then divide it by 1.24 = 1.30X10^-5???

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asked by Lloyd
  1. HCN is an acid. Treat it as an acid.
    HCN==> H^+ + CN^-

    Ka = (H^+)(CN^-)/(HCN) = 6.2 x 10^-10

    Now do the ICE table.
    I = initial concns:
    C = change in concns:
    E = equilibrium concns:

    Initial (before any ionization takes place):
    (HCN) = 1.24
    (H^+) = 0
    (CN^-) = 0

    change:
    (H^+) = +y
    (CN^-) = +y
    (HCN) = -y

    equilibrium:
    (HCN) = 1.24 - y
    (H^+) = 0 + y = y
    (CN^-) = 0 + y = y

    Substitute the equilibrium concns into the Ka expression and solve for y = (H^+), then pH = -log(H^+).

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