Thursday, March 23, 2023

# How To Solve For Concentration In Chemistry

## How To Calculate Volume Percent Concentration Of A Solution

Dilution Problems, Chemistry, Molarity & Concentration Examples, Formula & Equations

Volume percent is the volume of solute per volume of solution. This unit is used when mixing together volumes of two solutions to prepare a new solution. When you mix solutions, the volumes aren’t always additive, so volume percent is a good way to express concentration. The solute is the liquid present in a smaller amount, while the solute is the liquid present in a larger amount.

Calculate Volume Percent: volume of solute per volume of solution , multiplied by 100%

symbol: v/v %

v/v % = liters/liters x 100% or milliliters/milliliters x 100%

Example: What is the volume percent of ethanol if you dilute 5.0 milliliters of ethanol with water to obtain a 75-milliliter solution?

v/v % = 5.0 ml alcohol / 75 ml solution x 100% = 6.7% ethanol solution, by volume.

## Calculating Concentrations With Units And Dilutions

• Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

Calculating the concentration of a chemical solution is a basic skill all students of chemistry must develop early in their studies. What is concentration? Concentration refers to the amount of solute that is dissolved in a solvent. We normally think of a solute as a solid that is added to a solvent , but the solute could easily exist in another phase. For example, if we add a small amount of ethanol to water, then the ethanol is the solute, and the water is the solvent. If we add a smaller amount of water to a larger amount of ethanol, then the water could be the solute.

## To Find The Solubility Of Solutes

Wondering how to calculate molar solubility from $K_s_p$? Knowing the value of $K_s_p$ allows you to find the solubility of different solutes. Heres an example: The $K_s_p$ value of $Ag_2SO_4$ ,silver sulfate, is 1.4×$10^^5$. Determine the molar solubility.

First, we need to write out the dissociation equation: $K_s_p$=$^2$ $$Next, we plug in the K_s_p value to create an algebraic expression. 1.4×10^^5= ^2$$

1.4×$10^^5$= $4x^3$

$x$==1.5x$10^^2$ M

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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.

## More Ways To Calculate And Express Concentration

There are other easy ways to express the concentration of a chemical solution. Parts per million and parts per billion are used primarily for extremely dilute solutions.

g/L= grams per liter = mass of solute / volume of solution

F= formality = formula weight units per liter of solution

ppm = parts per million = ratio of parts of solute per 1 million parts of the solution

ppb= parts per billion = ratio of parts of solute per 1 billion parts of the solution.

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## What Is A Mole

A mole of anything, such as atoms, molecules, marbles or giraffes, is equal to 6.022 × 1023 individual instances of that thing. This happens to be the number of particles in exactly 12 grams of the most common form of the element carbon, number 6 on the periodic table of elements. The number of grams in 1 mole of a given element is under its symbol in its personalized box on the table.

To get the molar mass of a molecule, or the mass of 1 mol of those molecules, simply add the individual masses of the atoms, being sure to account for subscripts. So for a water molecule, H2O, you would add the mass of 1 mol O to the mass of 2 mol H to get about 18.015 g.

## 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 let’s 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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## Significance Of Solubility Product

Solubility depends on a number of parameters amongst which lattice enthalpy of salt and solvation enthalpy of ions in the solution are of most importance.

• When a salt is dissolved in a solvent the strong forces of attraction of solute must be overcome by the interactions between ions and the solvent.
• The solvation enthalpy of ions is always negative which means that energy is released during this process.
• The nature of the solvent determines the amount of energy released during solvation that is solvation enthalpy.
• Non-polar solvents have a small value of solvation enthalpy, meaning that this energy is not sufficient to overcome the lattice enthalpy.
• So the salts are not dissolved in non-polar solvents. Hence, for salt to be dissolved in a solvent, its solvation enthalpy should be greater than its lattice enthalpy.
• Solubility depends on temperature and it is different for every salt.

Salts are classified on the basis of their solubility in the following table.

 Category I

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## Converting Mass Concentrations To Ppm

Concentration Formula & Calculations | Chemical Calculations | Chemistry | Fuse School

Recall the definition of parts per million in mass of solute per volume of solution units derived above:

1 ppm = 1 g m-3 = 1 mg L-1 = 1 g mL-1

Some sample questions with worked solutions of converting w/v to ppm are given below.

Question 1.

What is its concentration in ppm?

Solution:

Step 1: Extract the data from the question

concentration = w/v = m/v = 1.25 g L-1

mass of solute = 1.25 g

volume of solution = 1 L

Step 2: Write the definition of ppm to be used

1 ppm = 1 g m-3 = 1 mg L-1 = 1 g mL-1

Step 3: Convert the mass of solute to required units

volume of solution is in litres so mass of solute must be in milligrams

mass = 1.25 g = 1.25 g × 1000 mg/g = 1250 mg

Step 4: Calculate concentration: divide mass by volume

concentration = mass ÷ volume

concentration = 1250 mg ÷ 1 L = 1250 mg L-1

Step 5: Write the concentration in ppm

1 ppm = 1 mg L-1 therefore 1250 mg L-1 = 1250 ppm

1.25 g L-1 = 1250 ppm

Question 2.

What is its concentration in ppm?

Solution:

Step 1: Extract the data from the question

concentration = w/v = m/v = 0.5 mg mL-1

mass of solute = 0.5 mg

volume of solution = 1 mL

Step 2: Write the definition of ppm to be used

1 ppm = 1 g m-3 = 1 mg L-1 = 1 g mL-1

Step 3: Convert the volume of solution to required units

mass of solute is in milligrams so volume of solution must be in litres

volume = 1 mL = 1 mL ÷ 1000 mL L-1 = 0.001 L

Step 4: Calculate concentration: divide mass by volume

ppm = mass ÷ volume

ppm = 0.5 mg ÷ 0.001 mL = 500 mg L-1

0.5 mg mL-1 = 500 ppm

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## Normality Problems And Examples

Question 1. In the following reaction calculate and find the normality when it is 1.0 M H3PO4

H3AsO4 + 2NaOH Na2HAsO4 + 2H2O

Solution:

If we look at the given reaction we can identify that only two of the H+ ions of H3AsO4 react with NaOH to form the product. Therefore, the two ions are 2 equivalents. In order to find the normality, we will apply the given formula.

N = Molarity × number of equivalents

N = 1.0 × 2

Therefore, normality of the solution = 2.0.

Question 2. Calculate the normality of the solution obtained by dissolving 0.321 g of the salt sodium carbonate in 250 mL water.

Solution:

Question 3. What is the normality of the following?

• 0.1381 M NaOH

a. N = 0.1381 mol/L × = 0.1381 eq/L = 0.1381 N

b. N = 0.0521 mol/L × = 0.156 eq/L = 0.156 N

Question 4. What will the concentration of citric acid be if 25.00 ml of the citric acid solution is titrated with 28.12 mL of 0.1718 N KOH?

Solution:

Na × =

Therefore, the concentration of citric acid = 0.1932 N.

Question 5. Find the normality of the base if 31.87 mL of the base is used in the standardization of 0.4258 g of KHP ?

Solution:

= 2.085 × 10-3 eq base/0.03187 L = 0.6542 N

Normality of the base is = 0.6542 N.

Question 6. Calculate the normality of acid if 21.18 mL is used to titrate 0.1369 g Na2CO3?

Solution:

= 2.583 × 10-3 eq acid/0.02118 L = 0.1212 N

Normality of the acid = 0.1212 N.

Try this:

## When To Use Normality

There are specific circumstances when its preferable to use normality rather than molarity or other unit of concentration of a chemical solution.

• Normality is used in acid-base chemistry to describe the concentration of hydronium and hydroxide . In this situation, 1/feq is an integer.
• The equivalence factor or normality is used in precipitation reactions to indicate the number of ions that will precipitate. Here, 1/feq is once again and integer value.
• In redox reactions, the equivalence factor indicates how many electrons can be donated or accepted by an oxidizing or reducing agent. For redox reactions, 1/feq may be a fraction.

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## How To Calculate The Final Concentration Of A Solution With Different Concentrations

When you mix two or more substances with different levels of concentration, the final solution does not simply equate to the combined concentration levels of the original ingredients. The nature of the experiment drives the ingredients used, including their individual concentration levels. Concentration levels typically represent a percent of the original ingredient by volume of the container, because there are no set units of concentration.

For example, if you mix 100 ml of a 10 percent concentration of compound A with 250 ml of a 20 percent concentration of the same compound, a mathematical formula involving the initial concentrations of the two solutions, as well as the volume of the final solution, allows you to work out the final concentration in percent of the volume of the new combined solution.

Determine the volume of each concentrated substance used in the experiment, by converting the concentration percentage to a decimal and then multiplying by the total volume of the solution. The calculation for the volume of compound A in the first concentration is x 100 ml, which is 10 ml. The calculation for the volume of compound A in the second concentration is x 250 ml, which is 50 ml.

Add these amounts together to find the total amount of compound A in the final mixture: 10 ml + 50 ml = 60 ml.

Add the two volumes together to determine the total volume of the final mixture: 100 ml + 250 ml = 350 ml.

## Using The Mass Per Volume Equation

• 1Find the mass of the solute mixed in with the solvent. The solute is the substance that youre mixing in to form your solution. If youre given the mass of the solute in your problem, write it down and be sure to label it with the correct units. If you need to find the mass of the solute, then weigh it on a lab scale and record the measurement.XResearch source
• If the solute youre using is a liquid, then you can also calculate the mass using the density formula, where density D = m/V, where m is the mass of the liquid and V is the volume. To find the mass, multiply the density of the liquid by the volume.
• Tip: If you need to use a scale, subtract the mass of the container youre using to hold the solute or else your calculations will be off.

• 2Record the total volume of the solution. The total volume of the solution is the amount of solvent plus the amount of solute added to it. If youre finding the volume in a lab, mix the solution in a graduated cylinder or beaker and look at the measurement. Measure the volume from the curve at the top of the solution, or the meniscus, to get the most accurate reading. Record the volume of the solution.XResearch source
• If you arent measuring the volume yourself, you may need to convert the mass of the solute into volume using the density formula.
• In our example for the concentration of 3.45 grams of salt in 2 liters of water, your equation would be C = / = 1.723 g/L.
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## Relating Solubilities To Solubility Constants

The of a solid is expressed as the concentration of the dissolved solid in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag+ concentration.

For example, let us denote the solubility of Ag2CrO4 as S mol L1. Then for a saturated solution, we have

• \

Substituting this into Eq 5b above,

thus the solubility is \.

Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas A2B and AB2, say, on the basis of their Ks values.

Note

It is meaningless to compare the solubilities of two salts having different formulas on the basis of their Ks values.

Example \

The solubility of CaF2 at 18°C is reported to be 1.6 mg per 100 mL of water. Calculate the value of Ksunder these conditions.

Solution

moles of solute in 100 mL S = 0.0016 g / 78.1 g/mol = \ mol

\^2 = ^2 = 4 × ^3 = 3.44 \times 10^\]

Example \

Estimate the solubility of La3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which Ks = 6.2 × 1012.

Solution

The equation for the dissolution is

If the solubility is S, then the equilibrium concentrations of the ions will be

= 3S = 2.08 × 105

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## How To Calculate Molality Of A Solution

Molality is used to express the concentration of a solution when you are performing experiments that involve temperature changes or are working with colligative properties. Note that with aqueous solutions at room temperature, the density of water is approximately 1 kg/L, so M and m are nearly the same.

Calculate Molality: moles solute per kilogram solvent

symbol: m

m = moles / kilogram

Example: What is the molality of a solution of 3 grams of KCl in 250 ml of water?

First, determine how many moles are present in 3 grams of KCl. Start by looking up the number of grams per mole of potassium and chlorine on a periodic table. Then add them together to get the grams per mole for KCl.

• K = 39.1 g/mol
• KCl = 39.1 + 35.5 = 74.6 g/mol

For 3 grams of KCl, the number of moles is:

* 3 grams = 3 / 74.6 = 0.040 moles

Express this as moles per kilogram solution. Now, you have 250 ml of water, which is about 250 g of water , but you also have 3 grams of solute, so the total mass of the solution is closer to 253 grams than 250. Using 2 significant figures, it’s the same thing. If you have more precise measurements, don’t forget to include the mass of solute in your calculation!

• 250 g = 0.25 kg
• m = 0.040 moles / 0.25 kg = 0.16 m KCl