# 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?

• Could you edit your question to show the chemical formula of sodium nitrate and calcium nitrate? That should help with figuring out the stoichiometry. – Karsten Theis May 29 at 13:17

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 &<=>> Na+ + NO3-}\label{rxn:R1}\tag{R1}\\ \ce{Ca(NO3)2 &<=>> 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-}) &= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\ &= \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}}\\ &= \pu{1.35 mol L-1} \end{align}