Question

Determine the equilibrium pH for a 1.0 mM solution of
NaHCO_{3} that is in equilibrium with a CO_{2(g)}
partial pressure of 0.10 atm. What is the pH after this solution
comes to equilibrium with the atmosphere (assume
H_{2}CO_{3}^{*} = 10^{-5.0}).
Please solve by spread sheet.

Answer #1

HCO_{3}^{-} (aq) + H_{2}O (l)
<----------> H_{2}O (l) + CO_{2} (g) +
OH^{-} (aq)

For this reaction, K_{b} = K_{w} / K_{a}
= 1.0*10^{-14} / 10^{-5} = 10^{-9}

Thus from the given data -

K_{b} = [OH^{-}] P_{CO2} /
[HCO_{3}^{-}]

10^{-9} = [OH^{-}] (0.10) / 0.001

[OH^{-}] = 10^{-11} M

Thus - [H^{+}] = 10^{-14} / 10^{-11} =
0.001 M

pH = -log [H^{+}] = **3**

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