Calculate acid ionization constant.?
1 Answer
Mar 3, 2018
Well, you should recall that
x = ["H"^(+)] = 10^(-"pH") = 1.778 xx 10^(-4) "M"
for the reaction
"HA"(aq) rightleftharpoons "H"^(+)(aq) + "A"^(-)(aq) .
Constructing an ICE table would yield
K_a = ((0 + ["H"^(+)]_(eq))(0 + ["A"^(-)]_(eq)))/(["HA"]_i - ["HA"]_"lost")
But the
color(blue)(K_a) = x^2/("0.80 M" - x)
= (1.778 xx 10^(-4))^2/(0.80 - 1.778 xx 10^(-4))
= color(blue)(3.95 xx 10^(-8))
Although keep in mind you only gave two sig figs.