Sure, let's break it down step-by-step to understand how the total molar concentration of ions in a [tex]$0.750 M$[/tex] solution of potassium carbonate ([tex]$K_2CO_3$[/tex]) can be calculated, assuming complete dissociation.
1. Dissociation Equation:
The dissociation of potassium carbonate in water can be represented as:
[tex]\[
K_2CO_3 (s) \rightarrow 2K^+ (aq) + CO_3^{2-} (aq)
\][/tex]
This means that one mole of [tex]$K_2CO_3$[/tex] dissociates into two moles of [tex]$K^+$[/tex] (potassium ions) and one mole of [tex]$CO_3^{2-}$[/tex] (carbonate ions).
2. Initial Molar Concentration:
The molar concentration of the [tex]$K_2CO_3$[/tex] solution is given as [tex]$0.750 M$[/tex]. This means that there are [tex]$0.750$[/tex] moles of [tex]$K_2CO_3$[/tex] in one liter of solution.
3. Moles of Ions Produced:
- For each mole of [tex]$K_2CO_3$[/tex], we get:
- [tex]$2$[/tex] moles of [tex]$K^+$[/tex] ions
- [tex]$1$[/tex] mole of [tex]$CO_3^{2-}$[/tex] ions
- Therefore, in a [tex]$0.750 M$[/tex] solution:
- The moles of [tex]$K^+$[/tex] ions produced will be [tex]$2 \times 0.750 = 1.5$[/tex] moles.
- The moles of [tex]$CO_3^{2-}$[/tex] ions produced will be [tex]$0.750$[/tex] moles.
4. Total Moles of Ions:
To find the total moles of ions, we sum the moles of [tex]$K^+$[/tex] and [tex]$CO_3^{2-}$[/tex] ions:
[tex]\[
\text{Total moles of ions} = 1.5 \text{ moles of } K^+ + 0.750 \text{ moles of } CO_3^{2-} = 2.25 \text{ moles}
\][/tex]
5. Total Molar Concentration of Ions:
Since the initial molar concentration of [tex]$K_2CO_3$[/tex] is [tex]$0.750 M$[/tex], the total molar concentration of ions in the solution will be:
[tex]\[
\text{Total molar concentration of ions} = 2.25 M
\][/tex]
So, in a [tex]$0.750 M$[/tex] solution of [tex]$K_2CO_3$[/tex], assuming complete dissociation, the total molar concentration of ions is [tex]$2.25 M$[/tex].
