Chapter 18, Problem 18.27P

Interpretation Introduction

Interpretation:

Preparation of standards of ammonia nitrogen solution of exactly $\text{1 ppm}$, $\text{2 ppm}$, $\text{4 ppm}$, $\text{8 ppm}$ concentration has to be explained.

Concept Introduction:

Dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$

Here,

$M_1$ denotes molarity of concentrated solution.

$V_1$ denotes volume of concentrated solution.

$M_2$ denotes molarity of diluted solution.

$V_2$ denotes volume of diluted solution.

Explanation of Solution

Since $1{\text{ mol NH}}_3$ has $1\text{ mol N}$, and mass of $1\text{ mol N}$ is $14.007\text{ g/mol}$, thus mass of  $\text{N}$ in $\text{28}{\text{.6 \% NH}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Concentration of N}=\left(\frac{28.6{\text{ g NH}}_3}{100\text{ mL}}\right)\left(\frac{14.007\text{ g/mol N}}{17.007{\text{ g/mol NH}}_3}\right)\left(\frac{{10}^6\text{ μg N}}{1{\text{ mL NH}}_3}\right)\\ =2.35\times {10}^5\text{ ppm}\end{array}$

Since it is a huge concentration, take $\text{1 mL}$ and dilute it to $1000\text{ mL}$. Concentration of stock thus obtained is calculated from Dilution formula as follows:

  $M_1{V}_1=M_2{V}_2$        (1)

Substitute $2.35\times {10}^5\text{ ppm}$ for $M_1$, $\text{1 mL}$ for $V_2$ , $1000\text{ mL}$ for $V_1$, in equation (1).

  $\left(2.35\times {10}^5\text{ ppm}\right)\left(1\text{ mL}\right)=M_2\left(\text{1000 mL}\right)$        (2)

Rearrange equation (2) to calculate value of $M_2$.

  $\begin{array}{c}{M}_2=\frac{\left(2.35\times {10}^5\text{ ppm}\right)\left(1\text{ mL}\right)}{\left(\text{1000 mL}\right)}\\ =235\text{ ppm}\end{array}$

For $\text{1 ppm}$ standard solution with $235\text{ ppm}$ stock solution,

Substitute $235\text{ ppm}$ for $M_1$, $\text{1 ppm}$ for $V_2$ , $1000\text{ mL}$ for $V_1$, in equation (1).

  $\left(\text{235 ppm}\right)V_1=\left(1\text{ ppm}\right)\left(\text{1000 mL}\right)$        (3)

Rearrange equation (3) to calculate value of $M_2$.

  $\begin{array}{c}{M}_2=\frac{\left(1\text{ ppm}\right)\left(\text{1000 mL}\right)}{\left(235\text{ ppm}\right)}\\ \approx 4.25\text{ mL}\end{array}$

Assume $\text{1 mL}$ equivalent $\text{1 g}$, weigh $4.25\text{ g}$ solution of $\text{28}{\text{.6 \% NH}}_3$ to prepare you new stock. Mass of $\text{N}$ in this stock is calculated as follows:

  $\begin{array}{c}\text{Concentration of N}=\left(\text{4}\text{.25 mL}\right)\left(\frac{28.6\text{ g }}{100\text{ mL}}\right)\left(\frac{14.007\text{ g/mol N}}{17.007{\text{ g/mol NH}}_3}\right)\left(\frac{1\text{ mg N}}{{10}^{-3}\text{ g N}}\right)\\ =1.00\text{ mg N/mL}\end{array}$

Hence mass in $\text{ppm}$ is calculated as follows:

  $\begin{array}{c}\text{Concentration}=\left(\frac{1.00\text{ mg N}}{\text{1 mL}}\right)\left(\frac{{10}^{-3}\text{ g}}{1\text{ mg}}\right)\left(\frac{1\text{ μg}}{{10}^{-6}\text{ g}}\right)\\ =1000\text{ ppm}\end{array}$

For exact $\text{1 ppm}$ standard solution,

Substitute $\text{1000 ppm}$ for $M_1$, $\text{1 mL}$ for $V_1$ , $1000\text{ mL}$ for $V_2$, in equation (1).

  $\left(\text{1000 ppm}\right)\left(\text{1 mL}\right)=M_2\left(\text{1000 mL}\right)$        (4)

Rearrange equation (4) to calculate value of $M_2$.

  $\begin{array}{c}{M}_2=\frac{\left(1000\text{ ppm}\right)\left(\text{1 mL}\right)}{\left(\text{1000 mL}\right)}\\ =1\text{ ppm}\end{array}$

For exact $\text{2}\text{.0 ppm}$ standard solution,

Substitute $\text{1000 ppm}$ for $M_1$, $\text{2 mL}$ for $V_1$ , $1000\text{ mL}$ for $V_2$, in equation (1).

  $\left(\text{1000 ppm}\right)\left(\text{2 mL}\right)=M_2\left(\text{1000 mL}\right)$        (5)

Rearrange equation (5) to calculate value of $M_2$.

  $\begin{array}{c}{M}_2=\frac{\left(1000\text{ ppm}\right)\left(\text{2 mL}\right)}{\left(\text{1000 mL}\right)}\\ =2\text{ ppm}\end{array}$

For exact $\text{4}\text{.0 ppm}$ standard solution,

Substitute $\text{1000 ppm}$ for $M_1$, $\text{4 mL}$ for $V_1$ , $1000\text{ mL}$ for $V_2$, in equation (1).

  $\left(\text{1000 ppm}\right)\left(\text{2 mL}\right)=M_2\left(\text{1000 mL}\right)$        (6)

Rearrange equation (6) to calculate value of $M_2$.

  $\begin{array}{c}{M}_2=\frac{\left(1000\text{ ppm}\right)\left(\text{4 mL}\right)}{\left(\text{1000 mL}\right)}\\ =4\text{ ppm}\end{array}$

For exact $\text{4}\text{.0 ppm}$ standard solution,

Substitute $\text{1000 ppm}$ for $M_1$, $\text{8 mL}$ for $V_1$ , $1000\text{ mL}$ for $V_2$, in equation (1).

  $\left(\text{1000 ppm}\right)\left(\text{2 mL}\right)=M_2\left(\text{1000 mL}\right)$        (7)

Rearrange equation (7) to calculate value of $M_2$.

  $\begin{array}{c}{M}_2=\frac{\left(1000\text{ ppm}\right)\left(\text{8 mL}\right)}{\left(\text{1000 mL}\right)}\\ =8\text{ ppm}\end{array}$

Hence to prepare exactly $\text{1 ppm}$, $\text{2 ppm}$, $\text{4 ppm}$, and $\text{8 ppm}$ dilute exactly $\text{1 mL}$ , $\text{2 mL}$$\text{4 mL}$ and $\text{8 mL}$ of $\text{1000 ppm}$ stock solution and make volume upto the mark in $\text{1000 mL}$ volumetric flask.