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QuantumCurt

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## Homework Statement

Calculate the pH of a solution containing 5.0 g of [itex]NaHCo_3[/itex] (molar mass = 84.0 g) in 250. mL of H20. [itex]NaHCO_3[/itex] is amphiprotic. [itex]k_a= 4.7x10^{-11}[/itex], [itex]k_b=2.3x10^{-8}[/itex]

This is for a General Chemistry II class, for the record.

## The Attempt at a Solution

I started by calculating the molarity of the NaHCO3 in the solution.

[tex] 5.0g x \frac{1 mole}{84.0 g} x \frac{1}{.250 L}=.238M[/tex]

With that calculated, I wrote out a balanced equation.

Then I set up an ICE (initial, change, equilibrium) chart

[tex] NaHCO_3 + H_2 O→N_aH_2CO_3^++OH^-[/tex]

With an initial molarity of .238 for NaHCO3, and values of 0 for NaCO3+ and OH-, I wrote an equilibrium constant expression for Ka

[tex] \frac{x^2}{.238-x}=4.7x10^{-11}[/tex]

Solving that for x, I get [itex]x=3.35x10^{-6}[/itex]

And taking the -log, I get a pH of 5.47.

The answer is supposed to be a pH of 9.9. What am I missing here? Is this something to do with the amphiprotic nature of NaHCO3? Am I writing an incorrect balanced equation?