![\rm Kb=\dfrac{[L][OH^-]}{[LOH]}](/tpl/images/0950/9039/84d1f.png)
![\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](/tpl/images/0950/9039/7025c.png)
Determine the [OH−] of a solution that is 0.150 M in CO2−3.
Kb(CO32−)=1.8×10−4...
Answers: 1
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