Question

What is the molarity of chloride ions in a ${\text{0}}{\text{.230 mol NaCl/L}}$ solution?

Answer 1

Let us first try to understand what molarity is. One of the most popular measures used to quantify the concentration of a solution is molarity (M), which shows the number of moles of solute per liter of solution (moles/Liter). Molarity is a unit of measurement which is used to find the volume of a solvent or the amount of a solute. To compute the molarity of a solution, the number of moles of solute should be divided by the total liters of the solution produced.

Complete answer:
According to the question, we have ${\text{0}}{\text{.230 mol NaCl/L}}$.
This means that $0.230$ moles of ${\text{NaCl}}$ are present in $1$ litre solution.
In this solution, we must determine the molarity of chloride ions. For this, we need to know how the ions dissociate in the solution.
So, we have,
${\text{NaCl}} \to {\text{N}}{{\text{a}}^ + } + {\text{C}}{{\text{l}}^ - }$
The moles of chloride ion dissociated per mole of solute is given by
$\dfrac{{{\text{ion}}}}{{{\text{solute}}}}{\text{ = }}\dfrac{{{\text{mol C}}{{\text{l}}^{\text{ - }}}}}{{{\text{mol NaC}}{{\text{l}}^{\text{ - }}}}} = \dfrac{{{\text{1 mol C}}{{\text{l}}^{\text{ - }}}}}{{{\text{1 mol NaC}}{{\text{l}}^{\text{ - }}}}}$
Now, we can determine the molarity of chloride ions in the solution as follows.

$${\text{M C}}{{\text{l}}^{\text{ - }}}{\text{ = M NaCl}} \times \dfrac{{{\text{ion}}}}{{{\text{solute}}}}{\text{ratio}} \\\ {\text{ = }}\dfrac{{{\text{0}}{\text{.230mol NaCl}}}}{{\text{L}}} \times \dfrac{{{\text{1mol C}}{{\text{l}}^{\text{ - }}}}}{{{\text{1mol NaCl}}}} \\\ = {\text{0}}{\text{.230mol C}}{{\text{l}}^{\text{ - }}}{\text{/L}} \\\ {\text{ = 0}}{\text{.230 M C}}{{\text{l}}^{\text{ - }}} \\\$$

Thus, is the molarity of chloride ions in a ${\text{0}}{\text{.230 mol NaCl/L}}$ solution is ${\text{0}}{\text{.230 M C}}{{\text{l}}^{\text{ - }}}$.

Note:
It is important to note that the mole ratio between the dissolved substance and the cations and anions it forms in solution determines the ion concentration in solution. If a compound dissociates into cations and anions, the minimum concentration of each of those two products will be equal to the original molecule's concentration. ${\text{NaCl}}$ when dissolved in water dissociate into one mole of ${\text{N}}{{\text{a}}^ + }$ ions and one mole of ${\text{C}}{{\text{l}}^ - }$ ions. Hence, the mole ratio of ${\text{N}}{{\text{a}}^ + }$ to ${\text{NaCl}}$ is $1{\text{ :1}}$.