# Find H3O+ of weak acid salt solution which compose of solute that can gain and give protons

Find $[\ce{H3O+}]$ of 0.100 M $\ce{NaHCO3}$ ($K_{\ce{H2CO3},1}=4.45×10^{-7}$, $K_{\ce{H2CO3},2}=4.7×10^{-11}$)

My work is: $$\ce{HCO3- + H2O<=>H2CO3 + OH-}$$

Initial: $[\ce{HCO3-}]$ = 0.100 M, $[\ce{H2CO3}]$ = 0, $[\ce{OH-}]$ = 0
Change: $[\ce{HCO3-}]$ = -X, $[\ce{H2CO3}]$ = +X, $[\ce{OH-}]$ = +X
Equilibrium: $[\ce{HCO3-}]$ = 0.100 - X, $[\ce{H2CO3}]$ = X, $[\ce{OH-}]$= X

$K_{b2}=\frac{Kw}{K_{a1}} = 2.25×10^{-8}$
$K_{b2} < 10^{-3}: 0.100 - x = x$
$K_{b2} = \frac{x^2}{0.100}$
$X = [\ce{OH-}] = 4.47×10^{-5}$ so $[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$, $[\ce{H+}] = 2.11×10^{-10}$

And the second equation is $$\ce{HCO3- + H2O<=>CO2 + H3O+}$$ For this equation $K_{a2}=4.7×10^{-11}$
So when I compared $K$ value between both of these equation, I will get $K_{b2} >> K_{a2}$ (more than $10^3$ times). So I think increase $[\ce{H3O+}]$ which occur or $[\ce{CO2}]$ can be ignored. So my answer $[\ce{H3O+}]$ is equal to $2.11×10^{-10}$ M.

But the solution in my book is:
From the formula:
$[\ce{H3O+}]= \sqrt{\dfrac{K_1 K_2 × [\ce{NaHCO3}] + k_1 k_w}{[\ce{NaHCO3}] +K_1}}$

$[\ce{H3O+}]=4.6×10^{-9}$ (solution answer)

Do I approximately second equation false or $K$ value is false or another thing?

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• The two equilibrium equations you need to worry about are: $$\ce{H2CO3 + H2O <=> HCO3- + OH-}\ \ \ \ K_{a1}$$ $$\ce{HCO3- + H2O <=> CO3^{2-} + H3O+}\ \ \ \ K_{a2}$$ – Ben Norris Nov 29 '14 at 14:56