Monday, 27 July 2015
Thursday, 23 July 2015
KNOW HOW TO PREPARE STANDARD SOLUTIONS FOR CHEMISTRY
KNOW HOW TO PREPARE STANDARD SOLUTIONS FOR CHEMISTRY
Chemists make two common types of "standard solutions":
•Molar solutions
•Normal solutions
Both of these solutions are concentrations (or “strengths”) of a particular component
(solute) that is dissolved in a solvent.
Making a Molar solution usually involves fewer mental steps than does making a Normal solution. Below, I will describe both methods.
PART 1: MAKING MOLAR (M) SOLUTIONS
A 1 Molar solution (1M) contains 1 mole of solute dissolved in a solution totaling 1 liter. If you use water as the solvent, it must be distilled and deionized. Do not use tap water.
A mole is the molecular weight (MW) expressed in grams (sometimes referred to as the
‘gram molecular weight’ (gMW) of a chemical). Thus, 1 M = 1 gMW of solute per liter of solution.
Problem: How much sodium chloride is needed to make 1 liter of an aqueous 1 M
solution?
Answer:
First, we calculate the molecular weight (MW) of sodium chloride. Checking the Periodic Table of Elements, we find that the atomic weight of sodium (Na) is 23 and the atomic weight of chlorine (Cl) is 35.5.
Therefore, the molecular weight of sodium chloride (NaCl) is: Na (23) + Cl (35.5) = 58.5 grams/mole.
To make a 1M aqueous solution of NaCl, dissolve 58.5 grams of NaCl in some distilled deionized water (the exact amount of water is unimportant; just add enough water to the flask so that the NaCl dissolves). Then add more water to the flask until it totals 1 liter. You’re done.
Similarly, a 2M solution of sodium chloride contains 117 grams of the salt (2 × 58.5 grams), topped-off with water to the one liter mark.
Likewise, a 0.1M solution of sodium chloride contains 5.85 grams (58.5 grams/10) of the salt, topped-off with water to the one liter mark.
MAKING MOLAR SOLUTIONS FROM CONCENTRATED AQUEOUS ACIDS AND BASES
Making a standard molar solution from aqueous acids or bases is a bit more involved than making a standard molar solution from a solid chemical. This is because nearly all liquid acids, no matter how concentrated they are, are already diluted to some extent with water (for instance, pure HCl is a gas, not a liquid, and it is rarely sold in its pure form). Let’s use sulfuric acid as our first example.
Problem: Make one liter of a 1 Molar (1M) aqueous solution of H2SO4.
Answer:
The first step is to read the label on the bottle of the H2SO4 reagent.
The label will tell you it's molarity. Although there are a variety of concentrations of
acids, concentrated H2SO4 often comes from the factory at a 18.0 Molar concentration (Table 1). This means that there are 18 moles of H2SO4 in each liter of solution (note: do not rely on Table 1; always check the label on the bottle). You need to make a much more
diluted solution, so you will add one mole of the concentrated reagent to a fresh batch of water. Your task is to calculate how many milliliters of reagent contain one mole of the acid.
We know from reading the label on the bottle (“18.0 Molar”) that one liter of reagent contains 18.0 moles of H2SO4. This means that 1 ml of reagent contains 0.018 moles of
H2SO4. Therefore,
1 ml x ml
= .
0.018 moles 1 mole
Solving for x, we find that we need 55.6 ml of H2SO4 reagent.
Therefore, we slowly add 55.6 ml of the H2SO4 reagent to about 500 ml of distilled deionized water, and then we top it off with more water to exactly the "1-liter" mark on
the flask.
You have successfully made a 1 Molar H2SO4 solution. This procedure works similarly with aqueous bases.
Caution: Never add water into a large volume of concentrated acid! You risk creating an explosion! The rule is: “Acid into water = you’re doing what ya oughta.”
“Water into acid = you might get blasted!” Therefore, always add a smaller volume of acid into a larger volume of water.
Table 1
TYPICAL CONCENTRATIONS OF CONCENTRATED ACIDS AND BASES (as written on the labels of their containers)
ACID/BASE NAME WT% DENSITY (sp. gr) (g/ml) MOLARITY
Acetic acid
99.7%
1.05 g/ml
17.4
Ammonium hydroxide
(aqueous ammonia)
28%
0.89 g/ml
14.6
Hydrochloric acid
37%
1.18 g/ml
12.0
Nitric acid (HNO3)
70%
1.40 g/ml
15.6
Phosphoric acid 85% 1.69 g/ml 14.7
Sulfuric acid
96%
1.84 g/ml
18.0
PART 2: MAKING NORMAL SOLUTIONS (N)
Compared to making Molar solutions, making Normal solutions can be a bit confusing. Aqueous solutions of acids and bases are often described in terms of their normality rather than their molarity. In order to properly make a Normal solution, the worker must understand the difference between a pure reagent and a diluted reagent.
A "1 Normal" solution (1 N) contains 1 “gram equivalent weight” (gEW) of solute, topped-off to one liter of solution. The gram equivalent weight is equal to the solute’s molecular weight, expressed as grams, divided by the valence (n) of the solute:
Equivalent weight (EW) = molecular weight
n
After the equivalent weight (or millieqiuvalent weight) has been calculated, then the following equation is used:
Weight of solute
N =
milliequivalent weight of solute × Volume (in ml) of dilution
The equivalent weight (or milliequivalent weight) of a substance depends upon the type of reaction in which the substance is taking part. Some different types of chemical reactions, along with how to determine a solute’s equivalent weight for each reaction, are given below.
MAKING A NORMAL SOLUTION WITH SALTS Problem: Calculate the normality of a sodium chloride solution prepared by
dissolving 2.9216 grams of NaCl in water and then topping it off with more water to
a total volume of 500.0 ml.
Answer:
•Checking the Periodic Table of Elements, we find that the molecular weight of
NaCl is 58.44
•n = 1 (because there is room in the molecule for only one replaceable H+ ion). In other words, one hydrogen atom can replace the sodium atom in NaCl.
•Therefore, the equivalent weight of NaCl is: 58.44 or 58.44
1
•Therefore, the milliequivalent weight of NaCl is: 58.44 or 0.05844
•The normality (N) is: N = 2.9216 grams
1000
[0.05844 × 500 ml] N = 0.099
MAKING NORMAL SOLUTIONS WITH PURE (NON-AQUEOUS) ACIDS
The equivalent weight of an acid is its molecular weight, divided by the number of replaceable hydrogen atoms in the reaction.
To clarify this concept, we will consider the following acids:
Hydrochloric acid (HCl) has one replaceable hydrogen ion (H+). Sulfuric acid (H2SO4)
has two replaceable hydrogen ions (2H+). The valences of these acids are determined by
their respective replaceable hydrogen ions (Table 2).: Table 2
HCl n = 1
HNO3 n = 1
H2SO4 n = 2
HF n = 1
So, for pure HCl, its MW is 36.46, its EW is 36.46 and therefore a 1N solution would be
36.46 grams of the pure chemical per liter. Note that, in the case of HCl, a 1N solution has the same concentration as a 1M solution..
To make a 1N H2SO4 solution from pure H2SO4, its MW is 98.08, and its
EW is: 98.08
2
Therefore,
EW = 49.04 grams per liter (98.08/2) (or 49.04 grams per 1000 milliliters)
So, a 1 N solution would be 49.04 grams of the pure chemical per liter. To make Normal acid solutions from aqueous reagents, see p. 7.
MAKING NORMAL SOLUTIONS FROM PURE ALKALIS (BASES)
The equivalent weight of a base is defined as "Its molecular weight divided by the number of hydrogen ions that are required to neutralize the base".
To understand the valences of alkalis, consider the following examples::
The (OH)- ion in Sodium Hydroxide (NaOH) can be neutralized by one hydrogen ion.
- -
The (OH)2
ions in Calcium hydroxide (Ca(OH)2) can be neutralized by two hydrogen
ions . As was the case with acids, the valences (n) of these bases are determined by their
respective replaceable hydrogen ions (Table 3):
Table 3
NaOH n = 1
Ca(OH)2 n = 2
So, for NaOH, its MW is 40, its EW is 40, and therefore a 1N solution would be 40 grams of the pure chemical per liter of water. You will also note that, in the case of NaOH, a 1N solution is the same concentration as a 1M solution.
For Ca(OH)2, its MW is 74, its EW is 74/2 = 37 (because n = 2). Therefore, a 1 N aqueous solution of Ca(OH)2 is 37 grams of the pure chemical per liter of water. Of course, many acidic reagents and basic reagents come from the factory in a diluted
aqueous form. To make Normal basic solutions from aqueous reagents, see p. 7.
HOW TO DETERMINE THE EQUIVALENT WEIGHT OF A SOLUTE IN REDOX REACTIONS
The equivalent weight of a substance undergoing oxidation or reduction is equal to its molecular weight divided by the total number of electrons gained or lost per molecule in that particular reaction. All you have to do is examine the redox equation and count the number of electrons involved in the reaction.
equivalent weight = molecular weight of solute number of e- gained or lost
Example #1:
Fe+++ + e- ==> Fe++
Equivalent weight of Fe = 58.85
1
Therefore, EW = 58.85
Example #2:
Sn+4 + 2e- ==> Sn+2
Equivalent weight Sn = 118.69
2
Therefore, EW = 59.34
Example #3:
Potassium Dichromate (K2Cr2O7)
Cr2O - -
+ 14H+
+ 6e-
==> 2Cr+++
+ 7H2O
Equivalent weight K2Cr2O7 = 294.19
6
Therefore, EW = 49.03
PREPARATION OF NORMAL SOLUTIONS FROM CONCENTRATED AQUEOUS REAGENTS
To prepare a standard Normal solution from a concentrated aqueous laboratory reagent, knowledge of the specific gravity (in grams/ml) and the percentage composition (by weight) of the reagent are required. The following example will produce a solution of a known normality. Subsequent titration would be necessary to achieve a highly precise normality, however a discussion of titration techniques is beyond the scope of this document. Consult a chemistry text book on quantitative analysis for more details on how to do a titration.
Problem: Calculate the volume of concentrated aqueous sulfuric acid, having a specific gravity of 1.842 and containing 96.0% H2SO4 (by weight), required to
prepare 2.0 liters of 0.20 N H2SO4 .
Answer:
Consulting the Periodic Table of Elements, we find that the molecular weight of
H2SO4 is 98.08.
Equivalent weight = molecular weight
n
Since n = 2, the equivalent weight is: 98.08
= 49.04 g/eq
2
The weight of the H2SO4 reagent that is required is calculated as follows: Equation #1:
Weight H2SO4 needed = 0.20 eq × 2.0 liters × 49.04 g liter eq
Multiplying and canceling-out some units, we get:
Weight H2SO4 needed = 19.616 grams
However, we want to find the volume of concentrated H2SO4 that is needed, not its weight. The weight of H2SO4 is related to its current diluted volume by the following equation:
Equation #2:
V(ml) × 1.842 g × 0.96
Weight =
ml
The “0.96” refers to the weight percent of H2SO4 in the reagent, which is written on the bottle’s label. In this case, it is 96%. Simplifying the equation, we arrive at:
Weight = V(ml) × 1.76832 grams
ml
Since “weight” is common to both equations #1 and #2, we can combine them into one equation:
Equation #3:
19.616 grams = V(ml) × 1.76832 grams
ml
Solving for V(ml) we get:
V(ml) = 19.616 grams
× ml
1.76832 grams
Canceling-out some units and rounding off, we get: V(ml) = 11.1 ml
We slowly pour 11.1 ml of H2SO4 into 1 liter of distilled deionized water, and then top it off with more water until the total volume of the solution is 2 liters. We have
successfully created a 0.2 N solution.
Problem: Calculate the volume of concentrated aqueous HCl, having a density of
1.188 g/ml and containing 38% HCl by weight, needed to prepare 2 liters of a 0.20 N
hydrochloric acid solution.
Answer:
The molecular weight of HCl = 36.461
The valence (n) of hydrogen = 1
Equation #1
Weight of HCl needed = 0.20 eq/Liter × 2.0 Liters × 36.461 g/eq
Re-writing the equation:
0.20 eq × 2000 ml × 36.461 grams
Weight of HCl needed =
1000 ml eq
Weight = 14.5844 grams
The weight of HCl is related to its current diluted volume by the following equation: Equation #2
Weight = V(ml) × 1.188 grams × 0.38
ml
Combining equations #1 and #2 gives us equation #3: Equation #3
14.5844 grams = V(ml) × 1.188 grams × 0.38
ml
Solving for V(ml):
V(ml) = 14.5844 grams × ml
1.188 grams × 0.38
Canceling-out some units and rounding off, we arrive at our answer: V(ml) = 32.3 ml
Therefore, we slowly add 32.3 milliliters of concentrated HCl to, say, 500 milliliters of distilled deionized water (remember: always add a smaller volume of acid to a larger volume of water), and then we top it off with more water until we have a total volume of
2 liters.
SOME HELPFUL TIPS
•Always use a fume hood when handling highly concentrated acids and bases. Wear a plastic apron, plastic gloves, and eye goggles. This is particularly important when working with hydrofluoric acid.
•Highly basic (alkali) reagents can, over time, slowly leach silica from the glass container into the reagent. For some applications in chemistry, a trace amount of silica contamination isn’t a problem, but if you need to make highly precise chemical measurements, the dissolved silica can affect the reagent’s chemical properties.
If you need a reagent that can produce highly precise results in quantitative analysis, then
you should store all newly prepared strong basic solutions in labeled plastic containers.
•Highly acidic reagents can be stored in either glass containers or in plastic containers. The exception is hydrofluoric acid, which etches glass and weakens it. Store hydrofluoric acid in a labeled plastic container.
•It is advisable to store concentrated reagents that come straight from the supplier in their original containers, with their labels intact.
•When preparing Molar solutions and Normal solutions, students sometimes make the mistake of adding the solute to a set volume of solvent. Doing this will produce a solution of the wrong concentration. Example: When making one liter of a 1 M solution of NaCl, do not add 58.5 grams of NaCl to 1 liter of water.
Instead, the correct way to make the solution is to add 58.5 grams of NaCl into a
container and then top it off with water to a total volume of 1 liter.
Saturday, 4 July 2015
Mwanza water quality lab is now accredited
The Mwanza Zonal Water Quality Laboratory has been accredited in the scope of “Chemical Analyses”
of water in accordance with ISO/IEC 17025 by the SADC Accreditation Service (SADCAS).
The accreditation certificate handover ceremony was held on 15 May 2015 at St Gasper Hotel in Dodoma the capital city of Tanzania. In her welcome remarks to the ceremony, Mrs Nadhifa Kemikimba the Direc- tor Division of Water Quality Services under the Ministry of Water, noted that Mwanza Zonal Water Quality Laboratory is the first government laboratory to be accredited in Tanzania. She reminisced on the journey to accreditation which started in 2004 by participating in Proficiency Testing. In 2012 the Ministry engaged a consultant to do a gap analysis of the laboratories management systems following which all issues raised were addressed and an application for accreditation was submitted to SADCAS in April 2014. The initial assessment was carried out in October 2014 and accreditation granted on 30
March 2015 for 15 water testing parameters. She noted that the accreditation would assist in assuring the safety of water.
Speaking during the ceremony Engi- neer Mbogo Futakamba, the Perma- nent Secretary Ministry of Water noted that Mwanza Zonal Water Qual- ity Laboratory was the first out of 16 laboratories under the Division of Wa- ter Quality Services of the Ministry of Water to be accredited following a directive from Honourable Jakaya Kik- wete, the President of the Republic of Tanzania to improve the quality of services. He noted the benefits de- rived from accreditation particularly accuracy and reliability of results. “Accreditation assures public account- ability” he said. He advised that the Ministry had a program in place for the other 15 laboratories under the Division of Water Quality Services to be accredited of which 5 laboratories
should be accredited by 2018. As the Accounting Officer under the Ministry, Engineer Futakamba com- mitted to supporting the program. He also advised that the Government would amend regulations to require all water testing laboratories to be accredited. In concluding his remarks Engineer Futakamba congratulated the team for a job well done and thanked all those who had assisted in the process.
In his presentation on the “Economic perspective on water quality”, Mr Edson Msangula, an accreditation con- sultant, highlighted the need for all water testing laboratories to be accredited and the market opportu- nities for accredited testing services in the mining, textile, fishing and environmental sectors. He also emphasized the need and importance of continued support to accredited laboratories and called upon the Government to review Public procurement Act to allow for new and emerging needs and require- ments. He also called upon the Government to allow laboratories to retain earned income from services and to review service charges so as to ensure sustainability of accredited laboratories. He concluded his presentation with a quote for Charles Darwin “It is not the strongest of the species that survives, nor the most intelligent, but the one most responsive to change.”
nia informed delegates about the role of the NAFP and the achievement it had made in promoting ac- creditation and marketing SADCAS services in Tanzania. She noted that Tanzania had the 2nd highest number of accreditations by SADCAS and the 4th highest number of accreditation applications under process
Mrs Maureen Mutasa the SADCAS Chief Executive Officer said that Mwanza Zonal Water Quality Labora- tory was the 4th testing laboratory and 9th laboratory in Tanzania to be accredited by SADCAS. The Mwanza Zonal Water Quality Laboratory had been granted the unique accreditation numbers TEST-5
0011 indicating that Mwanza Zonal Water Quality Laboratory is now a SADCAS accredited testing labo- ratory for chemical analyses of water. The certificate was issued on 30 March 2015 and is valid for 5 years until 29 March 2020 during which period SADCAS will undertake surveillance to ensure continued com- pliance with accreditation requirements. Mrs Mutasa said that Mwanza Zonal Water Quality Laboratory underwent a joint assessment by SADCAS and SANAS and had been issued with 2 accreditation certifi- cates with the latter being internationally recognized. “This is meant to ensure the credibility of SADCAS certificate whilst at the same time benefiting from skills transfer from SANAS”, she said. She informed the delegates that SADCAS had made great strides towards international recognition of its certificates having successfully undergone a pre peer evaluation in June 2014 and now set to undergo a peer evaluation from 26 to 31 May 2015. In conclusion Mrs Mutasa congratulated Mwanza Zonal Water Qual- ity Laboratory for the achievement and encouraged the Laboratory to extend its scope of accreditation to the other methods that the laboratory undertakes and encouraged the other laboratories under the Division of Water Quality Services to seek accreditation from SADCAS. She then handed over the certifi- cate to Engineer Futakamba who in turn handed over the certificate to the Honourable Professor Ju- manne A Maghembe, the Minister of Water who then handed over the certificate to Mr Heri M Chisute, the Manager Mwanza Zonal Water Quality Laboratory.
achievement “You make us stand 9 feet tall. The icing to all the Minis- try’s successes over the past 3 years.” he said and urged the Laboratory to continue to comply with SADCAS requirements in order to maintain the accreditation. Water is fundamental to life and the environment. It plays a central role in both economic and social development activi- ties. It plays a pivotal role in poverty alleviation through enhancement of food security, domestic hygiene and the environment. He noted the importance of accreditation in assuring the safety and cleanness of water and thanked SADCAS for bringing accreditation to Tanzania. The Honourable Professor Maghembe prided in having the best staff in the public service, who deliver albeit being the least funded. He reiterated his Ministry’s commitment to supporting all the laboratories under his Ministry and encouraged Mwanza Zonal Water Quality Laboratory to publicize its accreditation for competitiveness. He assured all that the regulations will be amended to include mandatory testing of water by an accredited laboratory.
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