Chapter 15, Problem 35A

Interpretation Introduction

Interpretation: The concentration of the chlorine ion is to be calculated.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

Answer to Problem 35A

The chlorin ion concentration is $0.113\text{M}$ .

Explanation of Solution

Given information:

Water’s volume, $V=25.0\text{mL}$

Volume of silver nitrate, $V=27.\text{2mL}$

The molarity if silver nitrate is, $M=0.1\text{04M}$

Calculation:

The number of moles is given as,

  $\begin{array}{l}M=\frac{n}{V}\\ \Rightarrow n=MV\end{array}$

The number of moles of silver ion is,

  $\begin{array}{c}M=\frac{n}{V}\\ n=M\times V=\left(\text{0}\text{.104}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\right)\text{×}\left(\text{27}{\text{.2×10}}^{\text{-3}}\text{L}\right)\\ \text{=}\text{2}{\text{.83×10}}^{\text{-3}}\text{mol}\end{array}$

In the given balanced equation, the mole ratio of silver and chlorine ion is $1:1$ . So, there are $2.83\times {10}^{-3}\text{mol}$ chlorine ion are present in the $25.0\text{mL}{\text{H}}_{\text{2}}\text{O}$ sample.

Therefore, the concentration of chlorine ion is,

  $\begin{array}{c}M=\frac{n}{V}\\ =\frac{\text{2}{\text{.83×10}}^{\text{-3}}\text{mol}}{\text{25}\text{.0mL}{\text{×10}}^{\text{-3}}\text{L}}\\ \text{=}\text{0}\text{.113M}\end{array}$

Conclusion

The concentration of chlorine ion is $0.11\text{3M}$ .