Question

Determine the [OH−] of a solution that is 0.150 M in CO2−3. Kb(CO32−)=1.8×10−4

Answer 1

The [OH−] of a solution : 0.0052 M

Further explanation

Kb is the base ionization constant  

LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)  

$\rm Kb=\dfrac{[L][OH^-]}{[LOH]}$

Dissociation of weak base of CO₃²⁻ :

Reaction

CO₃²⁻ + H₂O⇒ HCO₃⁻ + OH⁻

use ICE method :

I : 0.15               0             0

C : x                   x             x

E : 0.15-x            x            x

$\tt Kb=\dfrac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}\\\\1.8\times 10^{-4}=\dfrac{x^2}{0.15-x\approx 0.15}\\\\x^2=1.8\times 10^{-4}\times 0.15\\\\x=0.0052$