Calculate acid ionization constant.?

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Answer 1 · 2018-03-03T20:48:32

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.