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Question: What is the concentration of nitrate ions if equal volumes of 0.1 M \[AgN{O_3}\]​ and 0.1 M \(NaCl\)...

What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3AgN{O_3}​ and 0.1 M NaClNaCl are mixed together?
A. 0.1 M
B. 0.2 M
C. 0.05 M
D. 0.25 M

Explanation

Solution

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 of  ions=  Moles  of  solute  Volume  of  solution  in  LMolarity{\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 AgNO3AgN{O_3} and NaClNaCl
AgNO3+NaCl→AgCl(s)+ Na++NO3−AgN{O_3} + NaCl \to AgCl\left( s \right) + {\text{ }}N{a^ + } + {NO_3}^-
So,NO3−{NO_3}^- is obtained by dissociation of 0.1M  AgNO30.1M\;AgN{O_3} ​
Let the volume of both the solutions be 1L1L
We know that,
For AgNO3AgN{O_3}
1 mole1{\text{ }}mole of AgNO3AgN{O_3} gives 1 mole1{\text{ }}mole of NO3−{NO_3}^- ions on dissociation.
So, 0.1 mole0.1{\text{ }}mole AgNO3AgN{O_3} will furnish 0.1 moleNO3−{NO_3}^- ions in solution.
0.1 mole0.1{\text{ }}mole of Ag+  A{g^{ + \;}} & 0.1 mole0.1{\text{ }}mole of NO3−{NO_3}^- is present in 1L1L of solution.
ForNaClNaCl
1 mole1{\text{ }}mole of NaClNaCl gives 1 mole1{\text{ }}mole ofCl−C{l^ - } ions on dissociation
0.1 mole0.1{\text{ }}mole NaClNaCl solution contains 0.1 mole0.1{\text{ }}mole of Cl−C{l^ - }ions
0.1 mole0.1{\text{ }}mole of Na+  N{a^{ + \;}} & 0.1 mole0.1{\text{ }}mole of Cl−C{l^ - } ions is present in 1L1 L of solution.
∴\therefore Total moles of NO3−{NO_3}^-​ = 0.10.1 then the total volume of the mixture after mixing =2L2L
Now, we find the ion molarity.
∴\therefore Molarity of NO3−{NO_3}^-​ ions =     moles  of  NO−3    Volume  of  solution  in  L          \;\;\dfrac{{moles\;of\;N{O^ - }_{3\;}\;}}{{Volume\;of\;solution\;in\;L}}\;\;\;\;\; = 0.12L=0.05M\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.