I am trying to find the pH of the following problem (Answer key is $\mathrm{pH}=10.77$)

Calculate the $\mathrm{pH}$ of a $\pu{100 mL}$ solution containing $\pu{40.0 g}$ of $\ce{Na2CO3}$ after $\pu{400 mL}$ of $\pu{1.00 M}$ $\ce{HNO3}$ has been added. Ignore the volume of water produced by the reaction

My approach: $$\ce{Na_2CO_3~(aq)~+~2HNO_3~(aq)\rightarrow CO_2~(g)~+~H_2O~+~2NaNO_3~(aq)}$$

I start with $\pu{0.377 mol}$ $\ce{Na2CO3}$ and $\pu{0.4 mol}$ $\ce{HNO3}$. Because of the 2:1 ratio, I use up all of the $\ce{HNO3}$, and create $\pu{0.4 mol}$ $\ce{NaNO3}$ and have $\pu{0.177 mol}$ of $\ce{Na2CO3}$ left over.

I am unsure how to go about setting up the ice table. Should I use that $$\ce{NaNO3 + H2O -> HNO3 + Na+ + OH-}$$ where I have the initial amount of $\ce{NaNO3}$?


To solve this problem, you need to know the pKas of carbonic acid.

The pKas are 6.35 and 10.33.

If you add 1 equivalent of acid to $\ce{Na_2CO_{3}}$, the product will be $\ce{NaHCO_{3}}$ (sodium bicarbonate).

In this case, you are going 0.4-.377 = 0.023 moles past the point of 1 equivalent.

I would model the final solution as 0.354 moles of $\ce{HCO_{3}^-}$ and 0.023 moles $\ce{H_2CO_{3}}$ , and use the pKa of 6.35 and the Henderson-Hasselbalch equation to calculate pH.

The proposed answer 10.77 is incorrect.


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