Question

What is the concentration of nitrate ions if equal volumes of 0.1 M \[AgN{O_3}\]​ and 0.1 M $NaCl$ are mixed together?
A. 0.1 M
B. 0.2 M
C. 0.05 M
D. 0.25 M

Answer 1

Hint:Molarity is a concentration in terms of moles per liter of solution. Because an ionic compound dissociates into its components cations and anions in solution, the key is identifying how many moles of ions are produced during dissolution.

Formula used: \[Molarity{\text{ }}of\;ions = \;\dfrac{{Moles\;of\;solute}}{{\;Volume\;of\;solution\;in\;L}}\]

Complete answer:
The concentration of ions in solution depends on the mole ratio between the dissolved substance and the cations and anions it forms in solution. So, if a compound that dissociates into cations and anions, the minimum concentration of each of those two products will be equal to the concentration of the original compound.
First we determine the molarity of the solute
Now, The reaction between \[AgN{O_3}\] and \[NaCl\]
\[AgN{O_3} + NaCl \to AgCl\left( s \right) + {\text{ }}N{a^ + } + {NO_3}^- \]
So,\[{NO_3}^-\] is obtained by dissociation of \[0.1M\;AgN{O_3}\] ​
Let the volume of both the solutions be \[1L\]
We know that,
For \[AgN{O_3}\]
\[1{\text{ }}mole\] of \[AgN{O_3}\] gives \[1{\text{ }}mole\] of \[{NO_3}^-\] ions on dissociation.
So, \[0.1{\text{ }}mole\] \[AgN{O_3}\] will furnish 0.1 mole\[{NO_3}^-\] ions in solution.
\[0.1{\text{ }}mole\] of \[A{g^{ + \;}}\] & \[0.1{\text{ }}mole\] of \[{NO_3}^-\] is present in \[1L\] of solution.
For\[NaCl\]
\[1{\text{ }}mole\] of \[NaCl\] gives \[1{\text{ }}mole\] of\[C{l^ - }\] ions on dissociation
\[0.1{\text{ }}mole\] \[NaCl\] solution contains \[0.1{\text{ }}mole\] of \[C{l^ - }\]ions
\[0.1{\text{ }}mole\] of \[N{a^{ + \;}}\] & \[0.1{\text{ }}mole\] of \[C{l^ - }\] ions is present in \[1 L\] of solution.
$\therefore$ Total moles of \[{NO_3}^-\]​ = \[0.1\] then the total volume of the mixture after mixing =\[2L\]
Now, we find the ion molarity.
$\therefore$ Molarity of \[{NO_3}^-\]​ ions = \[\;\;\dfrac{{moles\;of\;N{O^ - }_{3\;}\;}}{{Volume\;of\;solution\;in\;L}}\;\;\;\;\;\] = \[\dfrac{{0.1}}{{2L}} = 0.05M\]

Hence, the correct option is C. 0.05 M

Note:
Molarity changes as the no. of moles of solute changes. While this calculation is straightforward when an ionic compound completely dissolves in solution, it's a bit trickier when a substance is only partially soluble.