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 $\ce{NaNO3}$?