# 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 *conjugate acid* of carbonate anion. Once you do that, you should then relate back to the **base association constant**, at

#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