What is the final molar concentration (molarity) of the nitrate ions in the resulting solution? A chemist mixes 16.3 mL of 1.23 M sodium nitrate with 6.28 mL 0.823 M solution of calcium nitrate. What is the final molar concentration (molarity) of the nitrate ions in the resulting solution?


What I know: I know I can add the volumes to use it to find the final molarity. After I multiply both volume and molarity and get amount of substance, I multiply by the ratio to get the amount of product and use those amounts to calculate molarity. But I don't know how to isolate the amounts of nitrate from the overall amount of product. Can someone explain how I do that?
First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:

$$
\begin{align}
\ce{NaNO3 &amp;&lt;=&gt;&gt; Na+ + NO3-}\label{rxn:R1}\tag{R1}\\
\ce{Ca(NO3)2 &amp;&lt;=&gt;&gt; Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2}
\end{align}
$$

Both nitrates are indeed well soluble in water (equilibrium is shifted to the far right), and no poorly soluble side-product is formed, e.g. we are likely to have a true solution consisting of hydrated cations $\ce{Na+}$, $\ce{Ca^2+}$ as well as $\ce{NO3-}$ anions.

As you already suggested, one can indeed use total amounts of substances $n_i$ and the preservation of volumes $V_i$ to find the final molar concentration on nitrate (here indices $1$ and $2$ refer to initial sodium and calcium nitrate solutions, respectively):

$$c(\ce{NO3-}) = \frac{n_1(\ce{NO3-}) + n_2(\ce{NO3-})}{V_1 + V_2}\label{eqn:1}\tag{1}$$

According to the stoichiometry (see \eqref{rxn:R1} and \eqref{rxn:R2}):

$$n_1(\ce{NO3-}) = n(\ce{NaNO3}) = c(\ce{NaNO3})\cdot V_1\tag{2.1}$$

$$n_2(\ce{NO3-}) = 2\cdot n(\ce{Ca(NO3)2}) = 2\cdot c(\ce{Ca(NO3)2})\cdot V_2\tag{2.2}$$

Finally, we can plug all the known quantities back to \eqref{eqn:1}:

$$
\begin{align}
c(\ce{NO3-}) &amp;= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\
  &amp;= \frac{\pu{1.23 mol L-1}\cdot\pu{16.3 mL} + 2\cdot\pu{0.823 mol L-1}\cdot\pu{6.28 mL}}{\pu{16.3 mL} + \pu{6.28 mL}}\\
  &amp;= \pu{1.35 mol L-1}
\end{align}
$$