# What concentration of NO_3- results when 697 mL of .515 M NaNO_3 is mixed with 635 mL of .785 M Ca(NO_3)_2?

Dec 4, 2015

There is a volume of $697$ $+$ $635$ $m L$. There are $0.857 \cdot m o l$ of nitrate ion.

#### Explanation:

Moles of $N a N {O}_{3}$ $=$ $0.697 \cdot L \times 0.515 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.359 \cdot m o l .$

Moles of $C a {\left(N {O}_{3}\right)}_{2}$ $=$ $0.635 \cdot L \times 0.785 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.498 \cdot m o l .$ Therefore there are $0.996$ $m o l$ nitrate from the calcium salt. (Why? Because each mole of calcium nitrate contributes 2 mol nitrate anion.)

In total there are thus $0.857$ $m o l$ nitrate ion dissolved in $1.332 \cdot L$ of solution. I have assumed that the volumes of solution are additive, which is entirely reasonable.

$\left[N {O}_{3}^{-}\right]$ $=$ $\frac{0.857 \cdot m o l}{1.332 \cdot L}$ $=$ ?? mol*L^-1?