1.2: Chemistry and Math Review
Soil science, as with any science, involves collection and interpretation of data. In order to appropriately record and interpret data in this course, students must have a fundamental understanding of unit conversions and chemistry. This lab will review some of the major concepts that are essential for success in Soils.
Learning Objectives
- Review basic chemistry and math skills that will be used throughout the semester.
- Chemistry and Math for Soil Scientists Problem Set
Materials
- Chemistry and Math for Soil Scientists Problem Set
Recommended Reading
Review relevant chemistry concepts in a chemistry textbook of your choosing.
Prelab Assignment
Using a chemistry textbook, and the conversion factors and formulas provided in this laboratory manual, consider the following questions.
- Define dimensional analysis. Describe how it can be useful for unit conversion, and how it can be used to check the accuracy of calculations.
- Define molarity and give several examples of how it can be expressed (labels or units).
- Describe, in general terms, the process of an acid-base titration.
- Note if any of the units listed in under Conversion Factors and Formulas are unfamiliar to you. If so, look that unit up and describe it in terms of units with which you are familiar.
- Are the units “Mg” and “mg” the same? If not, which one is larger?
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How many dimensions are there in the following units? Label each unit as a length, area, or volume.
- m:
- m 2 :
- m 3 :
Introduction
Dimensional Analysis
Quite often, our measurements are not in the same units in which we wish to express our results. However, converting from one measurement to another is not difficult with the correct conversion factors. The key is appreciating that different units can be used to express the same amount. Calling a sofa by another name like couch does not change anything about that piece of furniture, nor does expressing a person’s height in in instead of feet make them any taller or shorter. So by changing units, we do not change the amount, just the name we use to express that amount.
This is based on the principle that if we multiply any number by one, it does not change the number. For instance, we know that one hour is equal to 60 minutes, and that one minute is equal to 60 seconds. Consider the following fraction:
\[\frac{60\text{ minutes}}{60\text{ minutes}}\nonumber \]
We know that this fraction reduces to one because the top (the numerator) is equal to the bottom (the denominator). However, we know that 60 minutes is equal to one hour. So we could write the fraction:
\[\frac{60\text{ minutes}}{1\text{ hour}}\nonumber \]
While the numerical value of this fraction is 60, the amount of time represented in both the numerator and denominator are equal, and this fraction is, in a sense, equal to one. We use this concept, most likely without thinking about it, when we convert certain quantities in our head. If told that something will take about two hours, we can immediately convert this in our head to 120 minutes. Or if we need to be somewhere in half an hour, we can convert this to 30 minutes in our head without much effort. If we do think about it, however, we would recognize that the method used to make this conversion can be used to complete more complicated conversions that we cannot perform in our head. These conversions we performed in our head used the conversion factor 1 hour = 60 minutes and can be written as follows:
\[\frac{2\text{ hours}}{1}\times\frac{60\text{ minutes}}{1\text{ hour}} = 120\text{ minutes}\nonumber \]
In this case, when we write it out, the difference is that we flip the conversion factor so that 1 hour is in the numerator and 60 minutes is in the denominator. The most important aspect dimensional analysis is to always keep track of your units. By keeping track of your units, you make sure units cancel properly. In the previous example, the conversion factor has minutes in the denominator to cancel the minutes in the initial value; hours are in the numerator to take the place of the minutes.
\[\frac{120\text{ minutes}}{1}\times\frac{1\text{ hour}}{60\text{ minutes}}=2\text{ hours}\nonumber \]
If this is set up incorrectly, the conversion would yield:
\[\frac{120\text{ minutes}}{1}\times\frac{60\text{ minutes}}{1\text{ hour}}=\frac{7,200\text{ minutes}^2}{\text{hour}}\nonumber \]
This answer is not only incorrect, but it makes very little sense.
Chemical Titrations
Reactions in aqueous solution are commonly used to determine the unknown concentration of a dissolved substance. Titration is the technique used:
- A solution of known concentration (standard solution) is added quantitatively to a known volume of the solution of unknown concentration. For example, a burette is used to slowly add a solution with a known concentration of HCl to a specific volume of a solution with an unknown concentration of NaOH.
- The HCl and NaOH react to form NaCl and H 2 O. The point in the titration at which the last of the NaOH reacts with the added HCl is called the equivalence point . This point is commonly detected using an indicator that changes color at the equivalence point. Thus, you know to stop titrating when the color changes.
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Knowing the concentration and volume of the added HCl solution and the original volume of the NaOH solution allows us to calculate the concentration of NaOH in the original solution.
- We can calculate the number of moles of HCl in the solution using the following:
\[\text{Moles HCl}=(\text{volume HCl added})\times(\text{molarity of HCl})\nonumber \]
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- We know that when HCl and NaOH react, the equations is: HCl + NaOH → NaCl + H 2 O. Thus, one mole of HCl reacts with one mole of NaOH. So if we know the number of moles of HCl required for the reaction, we can calculate the number of moles of NaOH that reacted.
- Note: One practical application of titration is in determining blood alcohol content. Blood plasma is titrated with potassium dichromate, which oxidizes the alcohol. The amount of potassium dichromate required to do so is used to calculate the amount of alcohol present.
Let’s work an example:
We would like to determine the molarity of a NaOH solution of unknown concentration. Of this solution, 50 mL is titrated with 0.01 molar HCl solution. Of the HCl solution, 20 mL are required to reach the equivalence point (sometime called the end point ).
1. Calculate the moles of HCl used in the titration:
\[\text{Moles of HCl}=\frac{20\text{ mL HCl}}{1}\times\frac{0.01\text{ moles HCl}}{1\text{ L HCl}}\times\frac{1\text{ L HCl}}{1000\text{ mL HCl}}=2\times10^{-4}\text{ moles HCl}\nonumber \]
2. Calculate the moles of NaOH present:
Since we know that one mole of NaOH reacts with one mole of HCl, we know that 2 x 10 -4 moles of NaOH are present in the solution.
\[\text{Moles of NaOH}=\frac{2\times10^{-4}\text{ moles HCl}}{1}\times\frac{1\text{ mole NaOH}}{1\text{ mole HCl}}=2\times10^{-4}\text{ moles NaOH}\nonumber \]
3. Calculate the molarity of the NaOH:
The original volume of the NaOH solution was 50 mL. So 50 mL of solution contains 2 x 10 -4 moles of NaOH. Therefore, the molarity of NaOH is:
\[\frac{2\times10^{-4}\text{ moles NaOH}}{50\text{ mL NaOH}}\times\frac{1000\text{ mL NaOH}}{1\text{ L NaOH}}=\frac{0.004\text{ moles NaOH}}{\text{L}}\text{ or }0.004\text{ M NaOH}\nonumber \]
Let’s work another example:
We would like to determine the molarity of a Ca(OH) 2 solution of unknown concentration. Of this solution, 80 mL is titrated with 0.01 molar HCl solution. Of the HCl solution, 35 mL are required to reach the equivalence point. First, let’s write the equation for this reaction:
\[\ce{2HCl + Ca(OH)2 -> CaCl2 + 2H2O}\]
1. Calculate the moles of HCl used in the titration.
\[\text{Moles of HCl}=\frac{35\text{ mL HCl}}{1}\times\frac{0.01\text{ moles HCl}}{1\text{ L HCl}}\times\frac{1\text{ L HCl}}{1000\text{ mL HCl}}=3.5\times10^{-4}\text{ moles HCl}\nonumber \]
2. Calculate the moles of Ca(OH) 2 present:
In contrast to the reaction between HCl and NaOH, we see that two moles of HCl are required to react with one mole of Ca(OH) 2 . Therefore, the moles of Ca(OH) 2 in solution are:
\[\frac{3.5\times10^{-4}\text{ moles HCl}}{1}\times\frac{1\text{ mole Ca(OH)}_2}{2\text{ moles HCl}}=1.75\times10^{-4}\text{ moles Ca(OH)}_2\nonumber \]
- Calculate molarity of the Ca(OH) 2
The original volume of the Ca(OH) 2 solution was 80 mL. So 80 mL of solution contains 1.75 x 10-4 moles Ca(OH) 2 . Therefore, the molarity is
\[\frac{1.75\times10^{-4}\text{ moles Ca(OH)}_2}{80\text{ mL Ca(OH)}_2}\times\frac{1000\text{ mL Ca(OH)}_2}{1\text{ L Ca(OH)}_2}=\frac{2.19\times10^{-3}\text{ moles Ca(OH)}_2}{\text{L}}\text{ or }2.19\times10^{-3}\text{ M Ca(OH)}_2\nonumber \]
Activity 1 & Assignment: Problem Set
A problem set will be provided to you at the beginning of the lab. You will work on the problem set during the laboratory period, and turn it in as instructed.