S Per Million And Parts Per Billion
Two other concentration units are parts per million and parts per billion. These units are used for very small concentrations of solute such as the amount of lead in drinking water. Understanding these two units is much easier if you consider a percentage as parts per hundred. Remember that \ is the equivalent of 85 out of a hundred. A solution that is \ is 15 parts solute per 1 million parts solution. A \ solution is 22 parts solute per billion parts solution. While there are several ways of expressing two units of \ and \, we will treat them as \ or \ of solutes per \ solution, respectively.
For example, \ could be written as \ while \ can be written as \.
Problem : Earth’s Atmosphere
Statement: Earth’s atmosphere is a mixture of various gases. It extends up to an altitude of 90 km. Oxygen and nitrogen are dominant gases in the atmosphere. The mole percentage of oxygen and nitrogen is 21 % and 79 %. Find the mass concentration of each gas if the molar mass of oxygen and nitrogen is 32 g mol1 and 28 g mol1?
Solution: In 100 mol of air, the amount of oxygen and nitrogen is 21 mol and 79 mol respectively.
Calculating the mass of oxygen and nitrogen in 100 mol of air,
According to the ideal gas law, 1 mol of air occupies 22.4 dm3 at STP . For 100 mol of air, the volume will be 2240 dm3. Thus, V = 2240 dm3 = 2240 L.
The mass concentration of oxygen and nitrogen are calculated as follows:
If you add both concentrations, you get the density of air at STP.
How To Calculate Mole Fraction Of A Solution
Mole fraction or molar fraction is the number of moles of one component of a solution divided by the total number of moles of all chemical species. The sum of all mole fractions adds up to 1. Note that moles cancel out when calculating mole fraction, so it is a unitless value. Note some people express mole fraction as a percent . When this is done, the mole fraction is multiplied by 100%.
symbol: X or the lower-case Greek letter chi, , which is often written as a subscript
Calculate Mole Fraction: XA= /
Example: Determine the mole fraction of NaCl in a solution in which 0.10 moles of the salt is dissolved in 100 grams of water.
The moles of NaCl is provided, but you still need the number of moles of water, H2O. Start by calculating the number of moles in one gram of water, using periodic table data for hydrogen and oxygen:
- H = 1.01 g/mol
- O = 16.00 g/mol
- H2O = 2 + 16 = 18 g/mol
Use this value to convert the total number of grams of water into moles:
* 100 g = 5.56 moles of water
Now you have the information needed to calculate mole fraction.
- Xsalt= moles salt /
- Xsalt= 0.10 mol /
- Xsalt= 0.02
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Ion Concentrations In Solution
In Example \, the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL were calculated to be 1.43 M. Lets consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH4+ ions and one Cr2O72 ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72 ions:
Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72 anions and 2 mol of NH4+ cations ).
When carrying out a chemical reaction using a solution of a salt such as ammonium dichromate, it is important to know the concentration of each ion present in the solution. If a solution contains 1.43 M 2Cr2O7, then the concentration of Cr2O72 must also be 1.43 M because there is one Cr2O72 ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of 2Cr2O7 produces three ions when dissolved in water , the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.
Example \
What are the concentrations of all species derived from the solutes in these aqueous solutions?
Strategy:
A Classify each compound as either a strong electrolyte or a nonelectrolyte.
Solution:
Exercise \
What are the concentrations of all species derived from the solutes in these aqueous solutions?
- Answer a
Plot Of Concentration Versus Time For A Zero
Recall that the rate of a chemical reaction is defined in terms of the change in concentration of a reactant per change in time. This can be expressed as follows:
\text = -\frac}} = \text
=-\text
This is the integrated rate law for a zero-order reaction. Note that this equation has the form \text=\text. Therefore, a plot of versus t will always yield a straight line with a slope of -\text.
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The Common Ion Effect
It has long been known that the solubility of a sparingly soluble ionic substance is markedly decreased in a solution of another ionic compound when the two substances have an ion in common. This is just what would be expected on the basis of the Le Châtelier Principle whenever the process
is in equilibrium, addition of more fluoride ion will shift the composition to the left, reducing the concentration of Ca2+, and thus effectively reducing the solubility of the solid. We can express this quantitatively by noting that the solubility product expression
\^2 = 1.7 \times 10^ \label\]
must always hold, even if some of the ionic species involved come from sources other than CaF2. For example, if some quantity x of fluoride ion is added to a solution initially in equilibrium with solid CaF2, we have
- \
\^2 = S ^2 . \label\]
The plots shown below illustrate the common ion effect for silver chromate as the chromate ion concentration is increased by addition of a soluble chromate such as Na2CrO4.
Whats different about the plot on the right? If you look carefully at the scales, you will see that this one is plotted logarithmically Notice how a much wider a range of values can display on a logarithmic plot. The point of showing this pair of plots is to illustrate the great utility of log-concentration plots in equilibrium calculations in which simple approximations can yield straight-lines within the range of values for which the approximation is valid.
Example \
How To Calculate Concentration In Ppm
When youre working with solutions in chemistry, expressing the concentration of one component relative to the whole mixture is essential.
A lot of the time, youll be able to use more everyday measures like percent to convey the strength of the mixture, but in other cases youll need less common units like parts per million or PPM.
This is basically what it sounds like: The number of parts of the thing youre interested in for every 1 million parts of the solution. But its important to know how to perform a basic PPM calculation if youre going to communicate with other scientists.
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Calculating Moles Of Solute
Calculate the moles of copper sulfate in 250.00 mL of 0.020 mol L-1 copper sulfate solution.
n = moles of solute = ? mol
Extract the data from the question: c = concentration of solution = 0.020 mol L-1 V = volume of solution = 250.00 mL = 250.00 ÷ 1000 = 250.00 x 10-3 L = 0.25 L
Write the equation:
n = 0.020 × 0.25 = 0.0050 mol
What Is Meant By Concentration Of A Solution
In an aqueous solution, two parts exist, namely solute and solvent. They are the two basic solution concentration terms that you need to know. We always need to keep an account of the amount of solute in the solution. The amount of solute in the solvent is what is called the concentration of a solution. In chemistry, we define concentration of solution as the amount of solute in a solvent. When a solution has more solute in it, we call it a concentrated solution. Whereas when the solution has more solvent in it, we call it a dilute solution. Now that you understand the concept of what is concentration of solution lets move on to the different methods of expressing concentration.
The image shows a solution from the most dilute solution to the most concentrated solution.
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Using The Method Of Initial Rates To Determine Reaction Order Experimentally
2\ \text_\text_\rightarrow 4\ \text_+\text_
The balanced chemical equation for the decomposition of dinitrogen pentoxide is given above. Since there is only one reactant, the rate law for this reaction has the general form:
\text= \text^}
In order to determine the overall order of the reaction, we need to determine the value of the exponent m. To do this, we can measure an initial concentration of N2O5 in a flask, and record the rate at which the N2O5 decomposes. We can then run the reaction a second time, but with a different initial concentration of N2O5. We then measure the new rate at which the N2O5 decomposes. By comparing these rates, it is possible for us to find the order of the decomposition reaction.
S Per Million Calculations
Recall that, in general, concentration tells you how much solute is present in a solution.
concentration = amount of solute ÷amount of solution
A concentration in parts per million may refer to the mass of solute present in the volume of solution or it may refer to the mass of solute present in a mass of solution .
In SI units, w/w concentration would be given in kilograms of solute per kilograms of solution. So, a 1 part per million solution would be 1 kg of solute per 1 million kilograms of solution. And these masses are just too large to be useful in Chemistry laboratory. But we can divide the masses of solute and solution by 1 million to arrive at more useful units:
1 ppm |
ppm = mass of solute ÷ mass of solution
ppm = mass of solute ÷ mass of solution
You should practice rearranging the equations above in order to find mass of solute, volume of solution or mass of solution:
- To calculate mass of solute:
- To calculate volume of soution
- To calculate mass of soution
mass of solute = ppm × volume of solution
mass of solute = ppm × volume of solution
mass of solute = ppm × mass of solution
mass of solute = ppm × mass of solution
volume = mass of solute ÷ ppm
volume = mass of solute ÷ ppm
mass of solution = mass of solute ÷ ppm
mass of solution = mass of solute ÷ ppm
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How To Calculate Mass Percent Concentration Of A Solution
Mass percent composition is the easiest way to express the concentration of a solution because no unit conversions are required. Simply use a scale to measure the mass of the solute and the final solution and express the ratio as a percentage. Remember, the sum of all percentages of components in a solution must add up to 100%
Mass percent is used for all sorts of solutions but is particularly useful when dealing with mixtures of solids or anytime physical properties of the solution are more important than chemical properties.
Calculate Mass Percent: mass solute divided by mass final solution multiplied by 100%
symbol: %
Example: The alloy Nichrome consists of 75% nickel, 12% iron, 11% chromium, 2% manganese, by mass. If you have 250 grams of nichrome, how much iron do you have?
Because the concentration is a percent, you know a 100-gram sample would contain 12 grams of iron. You can set this up as an equation and solve for the unknown “x”:
12 g iron / 100 g sample = x g iron / 250 g sample
Cross-multiply and divide:
x= / 100 = 30 grams of iron
Ksp Chemistry: Complete Guide To The Solubility Constant
Are you learning chemistry but dont quite understand the solubility product constant or want to learn more about it? Not sure how to calculate molar solubility from $K_s_p$? The solubility constant, or $K_s_p$, is an important part of chemistry, particularly when youre working with solubility equations or analyzing the solubility of different solutes. When you have a solid grasp of $K_s_p$, those questions become much easier to answer!
In this $K_s_p$ chemistry guide, well explain the $K_s_p$ chemistry definition, how to solve for it , which factors affect it, and why its important. At the bottom of this guide, we also have a table with the $K_s_p$ values for a long list of substances to make it easy for you to find solubility constant values.
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How To Find The Concentration Of A Solution Using Different Methods
There are various methods of expressing concentration of a solution. You will usually see Chemists working with the number of moles. Pharmacists will use percentage concentrations instead of the number of moles. Hence it is important to understand all the methods of expressing concentration of solutions.
The concentration of solution formula is given as follows.
Cor S = \
We will also see other methods on how to calculate concentration of a solution based on the different methods of expressing concentrations.
Qhow To Calculate The Molar Concentration Of The Solution
The molar concentration unit is a conventionally widely used as concentration method. It is the number of moles of target substance dissolved in 1 liter of solution. Here is how to calculate the concentration.
x Ã· molecular weight
For example, let’s calculate the molar concentration of 2-mercaptoethanol . The necessary information is as follows.
- Specific gravity = 1.114 g/mL
- Purity = 100 w/w%
- Molecular weight = 78.13
In order to caluculate the concentration like above, it is necessary to know three points of “specific gravity “, “purity ” and “molecular weight”. The table below is a quick reference chart of common acid and base concentrations. In acid and alkali, there is a use for “neutralization titration”, “normality ” is often used.
ãQuick reference chart of common acid and base concentrationsã
Compound |
---|
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Calculating Mass Given Molality
We can also use molality to find the amount of a substance in a solution. For example, how much acetic acid, in mL, is needed to make a 3.0 m solution containing 25.0 g of KCN?
First, we must convert the sample of KCN from grams to moles:
\text = 25.0 \text \times = 0.38 \text
The moles of KCN can then be used to find the kg of acetic acid. We multiply the moles by the reciprocal of the given molality so that our units appropriately cancel. The result is the desired mass of acetic acid that we need to make our 3 m solution:
0.38 \text \times = 0.12\text
Once we have the mass of acetic acid in kg, we convert from kg to grams: 0.12 kg is equal to 120 g. Next, we use the density of acetic acid to convert to the requested volume in mL. We must multiply by the reciprocal of the density to accomplish this:
120.0 \text \times = 114.0 \text
Therefore, we require 114 mL of acetic acid to make a 3.0 m solution that contains 25.0 g of KCN.
Molarity vs. molality: In this lesson, you will learn how molarity and molality differ.
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If I Combine Two Solutions Of Known Concentrations And Volumes Can I Calculate The Concentration Of The New Solution By Adding The Two Concentrations Together And Dividing The Result By 2
Let’s use an example. Say, we have a #”100 mL”# If we wanted to find the final concentration of #HCl# , we would need to use the formula for molarity:
#”molarity” = “number of moles”/”volume of solution “#
For us, this would be:
#”molarity” = / = “2.56 M”#
Adding the concentrations together and then dividing the resulting value by #2# # -: 2 = 4.4 -: 2 = “2.2 M”# . This is incorrect, because the answer should have been #”2.56 M”# .
On a more conceptual level, this method doesnt work because it doesnt take the volumes into account.The initial volumes of the two solutions and the final volume of the mixed solution both affect the final concentration, but this method doesnt take that into account.
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Calculating Moles Given Molarity
To calculate the number of moles in a solution given the molarity, we multiply the molarity by total volume of the solution in liters.
How many moles of potassium chloride are in 4.0 L of a 0.65 M solution?
\text_}=\frac_}}}
0.65 \text = \frac_\text}}
\text_\text = = 2.6 \text
There are 2.6 moles of KCl in a 0.65 M solution that occupies 4.0 L.
Key Concepts And Summary
Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.
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