Chapter 6.5, Problem 6.16PP

Interpretation Introduction

Interpretation:

The concentration of sodium ion in $mol/L$ has to be calculated.

Concept Introduction:

The concentration of an ion in $eq/L$ can be calculated as follows,

    $\begin{array}{c}\text{eq/L}\text{=}\left(\frac{\text{eq}}{\text{mol}\text{ion}}\right)\text{×M}\\ \\ \text{=}\left(\frac{\text{eq}}{\text{mol}\text{ion}}\right)\text{×}\left(\frac{mol}{L}\right)\end{array}$

Where, $\frac{eq}{molion}$ is the number of charges on the ion irrespective of the sign and $M$ is the molarity of the solution.