What is the hydroxide-ion concentration?

Question

What is the hydroxide-ion concentration in a 0.13 M solution of Na2CO3? For carbonic acid, Ka1 = 4.2 × 10–7 and Ka2 = 4.8 × 10–11. (Kw = 1.0 × 10–14)

Answer 1

The detailed solution (no assumptions) involves solving a system of equations.  In this case there are six variables, and six equations are needed to obtain the value for each variable.  This is pretty involved, but here's enough to get started:

Let [OH^-]= a

Let [H⁺] = b

Let [Na⁺] = c

Let [H₂CO₃] = d

Let [HCO₃^-] = e

Let [CO₃^2-] = f

Here are the six equations:

Water equilibrium:

ab = 10^-14

Given:

c = 0.26

Mass balance:

d + e + f = 0.13

Charge balance:

b + c = a + e + 2f

Equilibrium expression:

4.2 x 10^-7 = be/d

Equilibrium expression:

4.8 x 10^-11 = bf/e

 

Now solve for "a" by starting with the charge balance equation.

 

If you want the simpler solution (using assumptions), here it is:

 

CO₃^2- + H₂O = HCO₃^- + OH^-,   K_b2 = 10^-14/K_a2 = 2.08 x 10^-4

HCO₃^- + H₂O = H₂CO₃ + OH^-,  K_b1 = 10^-14/K_a1 = 2.38 x 10^-8 

 

Let x = the molar concentration of CO₃^2- which reacts with water to form HCO₃^-

Let y = the molar concentration of the initially-formed HCO₃^- which further reacts with water to form H₂CO_3.

 

Now use the first equilibrium constant expression above to write the following:

 

2.08 x 10^-4 = ((x-y)(x+y)) / (.13 - x).

 

Because K_b1 is very small, we can assume that y is much smaller than x, and we can ignore it in the calculation:

 

2.08 x 10^-4 = x² / (.13 - x)

 

Now solve the quadratic equation above to find that x = 0.0051 M ≈ [OH^-]