Chapter 6, Problem 6.111QP

(a)

Interpretation Introduction

Interpretation:

The number of mole of $N{a}^{+}in1L$ in a solution containing $154meq/LofN{a}^{+}$ 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.

Answer to Problem 6.111QP

The number of mole of $N{a}^{+}in1L$ solution is $0.154mol/L.$

Explanation of Solution

$154meq/LofN{a}^{+}$ is equal to $0.154eq/LofN{a}^{+}.$

The number of mole of $N{a}^{+}in1L$ solution can be calculated as follows,

    $\begin{array}{c}\text{eq/L}\text{=}\left(\frac{\text{eq}}{\text{mol}\text{ion}}\right)\text{×M}\\ \\ \text{=}\left(\frac{\text{eq}}{\text{mol}\text{ion}}\right)\text{×}\left(\frac{mol}{L}\right)\\ \\ 0.154eq/L=\left(1eq/mol\right)\text{×}\left(\frac{mol}{L}\right)\\ \\ \left(\frac{mol}{L}\right)=\frac{0.154eq/L}{1eq/mol}\\ \\ =0.154mol/L.\end{array}$

The molarity of sodium ion is $0.154mol/L.$

(b)

Interpretation Introduction

Interpretation:

The number of mole of $C{l}^{-}in1L$ in a solution containing $154meq/LofC{l}^{-}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.111QP

The number of mole of $C{l}^{-}in1L$ solution is $0.154mol/L.$

Explanation of Solution

$154meq/LofC{l}^{-}$ is equal to $0.154eq/LofC{l}^{-}.$

The number of mole of $C{l}^{-}in1L$ solution can be calculated as follows,

    $\begin{array}{c}\text{eq/L}\text{=}\left(\frac{\text{eq}}{\text{mol}\text{ion}}\right)\text{×M}\\ \\ \text{=}\left(\frac{\text{eq}}{\text{mol}\text{ion}}\right)\text{×}\left(\frac{mol}{L}\right)\\ \\ 0.154eq/L=\left(1eq/mol\right)\text{×}\left(\frac{mol}{L}\right)\\ \\ \left(\frac{mol}{L}\right)=\frac{0.154eq/L}{1eq/mol}\\ \\ =0.154mol/L.\end{array}$

The molarity of chloride ion is $0.154mol/L.$