Answer:
Ka = 4.9 X 10^-9
Explanation:
when the pH is 4.95
the [H+] = 10^-4.95
[H+] = 1.122 e-5
since the equilibrium equation is:
HCN <=> 1 mole H+ & 1 mole CN-
then
[H+] = [CN-]
when
HCN <=> H+ & CN-
Ka = [ H+] [CN-] / [HCN]
Ka = [ 1.122 e-5] [ 1.122 e-5] / [0.255]
Ka = 1.256 e-10 / 0.255
Ka = 4.94 e-9
which rounds off to 2 sig figs into
Ka = 4.9 X 10^-9
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p.s. it was the pH pf 4.95 that limited the calculations to 2 sig figs
