Calculate acid ionization constant.?

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1 Answer
Mar 3, 2018

Well, you should recall that x = ["H"^(+)] in an acid dissociation. Therefore, you can find x first.

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 "HA" lost became stoichiometrically "H"^(+) and "A"^(-), so ["HA"]_"lost" = ["H"^(+)]_(eq) = ["A"^(-)]_(eq). Therefore,

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.