# Question #94bfd

Jan 21, 2017

$1.73 \cdot m o l \cdot {L}^{-} 1 \text{ with respect to sodium ion.}$

#### Explanation:

We sum up the the total number of moles of $N {a}^{+}$.

$\text{Moles}$ $=$ $\text{concentration}$ $\times$ $\text{volume}$

$= 25.0 \times {10}^{-} 3 L \times 1.15 \cdot m o l \cdot {L}^{-} 1 + 2 \times 25.0 \times {10}^{-} 3 L \times 1.15 \cdot m o l \cdot {L}^{-} 1 = 8.63 \times {10}^{-} 2 \cdot m o l$ with respect to sodium ion.

And this is dissolved in a volume of $50.0 \cdot m L$.

The new concentration is thus, $\frac{8.63 \times {10}^{-} 2 \cdot m o l}{50.0 \cdot m L} = 1.73 \cdot m o l \cdot {L}^{-} 1$ with respect to sodium ion.

Why did I multiply one of the starting concentrations by 2?