# What are the total number of ions in solution if a 25.0*g mass of calcium chloride is dissolved in a 500*mL volume of water?

Jul 20, 2017

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$

#### Explanation:

And so we calculate the molar quantity WITH RESPECT to $C a C {l}_{2}$, and take the quotient.......

$\frac{\frac{25.0 \cdot g}{110.98 \cdot g \cdot m o {l}^{-} 1}}{0.50 \cdot L} = \frac{0.225 \cdot m o l}{0.50 \cdot L} = 0.450 \cdot m o l \cdot {L}^{-} 1$, with respect to $C a C {l}_{2} \left(a q\right)$

But we know that calcium chloride speciates in solution to give THREE equivalents of ions......

$C a C {l}_{2} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} C {a}^{2 +} + 2 C {l}^{-}$

And thus $\left[C {a}^{2 +}\right] = 0.450 \cdot m o l \cdot {L}^{-} 1$....and by stoichiometry....

$\left[C {l}^{-}\right] = 2 \times \left[C {a}^{2 +}\right] = 0.900 \cdot m o l \cdot {L}^{-} 1$

Are you with me? The stoichiometry of the formula determines the respective concentration of the ions in solution. How would this change if we had for instance, $\left[N a C l \left(a q\right)\right] = 0.450 \cdot m o l \cdot {L}^{-} 1$; what are $\left[N {a}^{+}\right]$ and $\left[C {l}^{-}\right]$?