Question

Equal volumes of $0.1\mathrm{M}{\mathrm{AgNO}}_3$ and $0.2\mathrm{M}\mathrm{NaCl}$are mixed. The Concentration of ${{\mathrm{NO}}_3}^{-}$ ions in the mixture will be______

Answer 1

Step 1: Analyze the corresponding reaction.

One mol of Silver Nitrate (${\mathrm{AgNO}}_3$) reacts with one mol of Sodium Chloride ($\mathrm{NaCl}$) to form one mole of Silver chloride ($\mathrm{AgCl}$) and one mole of Sodium Nitrate (${\mathrm{NaNO}}_3$).

The reaction between Silver Nitrate and Sodium Chloride proceeds as follows:

$\underset{\mathrm{Silver}\mathrm{Nitrate}}{{\mathrm{AgNO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Sodium}\mathrm{Chloride}}{\mathrm{NaCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Silver}\mathrm{Chloride}}{\mathrm{AgCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{Nitrate}}{{\mathrm{NaNO}}_3\left(\mathrm{aq}\right)}$

Step 2: Find concentration of Nitrate ions in one mole of Silver Nitrate

According to the reaction, Silver Nitrate interacts entirely with Sodium Chloride at $0.1\mathrm{M}$.

Molarity is defined as the number of moles of solute per litre of solution

So, $0.1\mathrm{M}$ Silver Nitrate is$0.1\mathrm{mol}$Silver Nitrate in $1\mathrm{L}$ of solution.

Hence, in $1\mathrm{L}$ of solution, $1\mathrm{mol}$ of silver nitrate yields$0.1$moles of nitrate ions.

Step 3: Find concentration of nitrate ions in mixture.

When an equal volume of two solutions is combined, the volume doubles, and the concentration changes.

The volume of the solution after mixing $=2\mathrm{L}$

In a $2\mathrm{L}$ solution of nitrate, the molarity is:

$$\begin{gathered} =\frac{0.1\mathrm{mol}}{2\mathrm{L}} \\ =0.05\mathrm{M} \end{gathered}$$

Thus, the new nitrate ion concentration in the mixture will be $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}$.