# Question 2cabc

Apr 16, 2017

The number of ions present in a given amount of moles of a compound is calculated using Avogadro's number and the number of ions represented in the formula of its solution.

#### Explanation:

Example:
If we had one mole of $N a O H$ which is a strong base solid dissolved in $1 l$ water, we would have an aqueous solution of ions.

The number of particles in one mole of NaOH is $6.022 \times {10}^{23}$
The number of $N {a}^{+} \left(a q\right)$ ions in one mole of its solution is $6.022 \times {10}^{23}$
The number of OH^#-(aq) ions in one mole of its solution is $6.022 \times {10}^{23}$

Now it looks like there are twice as many total ions as there are particles of the original solid. And this agrees with the chemical equation resulting from the solution of the solid $N a O H$ in water.

$N a O H \left(s\right) \to N {a}^{+} \left(a q\right) + O {H}^{-}$(aq)

Another example:
If we dissolve one mole of fertilizer consisting of ammonium sulfate into a solution and add enough water to bring the volume up to $1 l$

${\left(N {H}_{4}\right)}^{2} S {O}_{4} \left(s\right) \to 2 N {H}_{4}^{+} \left(a q\right) + S {O}_{4}^{2 -}$(aq)

Note that in this solution we have three ions formed from each particle of the solid when in solution.

The number of particles in one mole of ${\left(N {H}_{4}\right)}^{2} S {O}_{4} \left(s\right)$ is $6.022 \times {10}^{23}$
The number of $N {H}_{4}^{+} \left(a q\right)$ ions in one mole of solution is $2 \times 6.022 \times {10}^{23} = 12.044 \times {10}^{23}$
The number of $S {O}_{4}^{2 -}$ ions in one mole of solution is $6.022 \times {10}^{23}$