Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 18, Problem 18.27P
Interpretation Introduction

Interpretation:

Preparation of standards of ammonia nitrogen solution of exactly 1 ppm, 2 ppm, 4 ppm, 8 ppm concentration has to be explained.

Concept Introduction:

Dilution formula is given as follows:

  M1V1=M2V2

Here,

M1 denotes molarity of concentrated solution.

V1 denotes volume of concentrated solution.

M2 denotes molarity of diluted solution.

V2 denotes volume of diluted solution.

Expert Solution & Answer
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Explanation of Solution

Since 1 mol NH3 has 1 mol N, and mass of 1 mol N is 14.007 g/mol, thus mass of  N in 28.6 % NH3 is calculated as follows:

  Concentration of N=(28.6 g NH3100 mL)(14.007 g/mol N17.007 g/mol NH3)(106 μg N1 mL NH3)=2.35×105 ppm

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

  M1V1=M2V2        (1)

Substitute 2.35×105 ppm for M1, 1 mL for V2 , 1000 mL for V1, in equation (1).

  (2.35×105 ppm)(1 mL)=M2(1000 mL)        (2)

Rearrange equation (2) to calculate value of M2.

  M2=(2.35×105 ppm)(1 mL)(1000 mL)=235 ppm

For 1 ppm standard solution with 235 ppm stock solution,

Substitute 235 ppm for M1, 1 ppm for V2 , 1000 mL for V1, in equation (1).

  (235 ppm)V1=(1 ppm)(1000 mL)        (3)

Rearrange equation (3) to calculate value of M2.

  M2=(1 ppm)(1000 mL)(235 ppm)4.25 mL

Assume 1 mL equivalent 1 g, weigh 4.25 g solution of 28.6 % NH3 to prepare you new stock. Mass of N in this stock is calculated as follows:

  Concentration of N=(4.25 mL)(28.6 g 100 mL)(14.007 g/mol N17.007 g/mol NH3)(1 mg N103 g N)=1.00 mg N/mL

Hence mass in ppm is calculated as follows:

  Concentration=(1.00 mg N1 mL)(103 g1 mg)(1 μg106 g)=1000 ppm

For exact 1 ppm standard solution,

Substitute 1000 ppm for M1, 1 mL for V1 , 1000 mL for V2, in equation (1).

  (1000 ppm)(1 mL)=M2(1000 mL)        (4)

Rearrange equation (4) to calculate value of M2.

  M2=(1000 ppm)(1 mL)(1000 mL)=1 ppm

For exact 2.0 ppm standard solution,

Substitute 1000 ppm for M1, 2 mL for V1 , 1000 mL for V2, in equation (1).

  (1000 ppm)(2 mL)=M2(1000 mL)        (5)

Rearrange equation (5) to calculate value of M2.

  M2=(1000 ppm)(2 mL)(1000 mL)=2 ppm

For exact 4.0 ppm standard solution,

Substitute 1000 ppm for M1, 4 mL for V1 , 1000 mL for V2, in equation (1).

  (1000 ppm)(2 mL)=M2(1000 mL)        (6)

Rearrange equation (6) to calculate value of M2.

  M2=(1000 ppm)(4 mL)(1000 mL)=4 ppm

For exact 4.0 ppm standard solution,

Substitute 1000 ppm for M1, 8 mL for V1 , 1000 mL for V2, in equation (1).

  (1000 ppm)(2 mL)=M2(1000 mL)        (7)

Rearrange equation (7) to calculate value of M2.

  M2=(1000 ppm)(8 mL)(1000 mL)=8 ppm

Hence to prepare exactly 1 ppm, 2 ppm, 4 ppm, and 8 ppm dilute exactly 1 mL , 2 mL4 mL and 8 mL of 1000 ppm stock solution and make volume upto the mark in 1000 mL volumetric flask.

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