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.