What is the equilibrium concentration of "OH"^(-) in a "0.120 M" solution of "Na"_2"CO"_3?
1 Answer
I got
Well, you have to be the one to look up the
K_aK_b = K_w = 10^(-14)
=> K_b = 10^(-14)/K_a
The
Now we simply write the association reaction in water (you'll always do it in water in general chemistry!), and if you wish, you can write an ICE table:
"CO"_3^(2-)(aq) " "+" ""H"_2"O"(l) rightleftharpoons "HCO"_3^(-)(aq) + "OH"^(-)(aq)
"I"" ""0.120 M"" "" "" "" "-" "" "" "" ""0 M"" "" "" ""0 M"
"C"" "-x" "" "" "" "" "-" "" "" "" "+x" "" "" "+x
"E"" "(0.120 - x)"M"" "-" "" "" "" "x" M"" "" "" "x" M"
And from this we obtain the equilibrium expression:
2.08 xx 10^(-4) = x^2/("0.120 M" - x) ,
with
Rearrange to get the quadratic form...
x^2 + 2.08 xx 10^(-4)x - (2.08 xx 10^(-4))(0.120) = 0
Solve using the quadratic formula or Wolfram Alpha to get:
color(blue)(ul(x = "0.0049 M" = ["OH"^(-)]))
What is the