Definition of Solution Concentration – To make it easier to understand what is the concentration of a solution, we should know about the meaning of the solution itself. Solution in chemistry has the meaning of a homogeneous mixture with a composition ratio according to its constituent components. One example of a solution in chemistry is H2SO4 (sulfuric acid). If a solution of H2SO4 (sulfuric acid) is electrified, it will conduct electricity.
In general, a solution consists of one type of solute and one solvent. Solvent (solvent) and solute (substance that is dissolved) are usually often heard and mentioned. Solvent is a component that physically does not change if a solution is formed, whereas all components present in the solute will dissolve in the solvent.
Even though the solution is a homogeneous mixture, the composition of each solution can vary. For example, there are two solutions in which each solvent contains one liter, but the amount of salt dissolved is different. From the two salt solutions just now, others couldn’t tell how much salt was contained therein.
Therefore, to find out information about the relative amount of solute and solvent present in the solution, the term concentration of the solution is used. The concentration of a solution is the amount of substance dissolved in each unit of solution or solvent. In simple terms, the concentration of a solution can provide an overview or information about the ratio of the amount of solute to the amount of solvent.
The concentrations of solutions commonly used in laboratories are Molarity, Molality, Normality, Mole Fraction, Concentration in Percent, Parts per Million (ppm) and Parts per Billion (ppb), and Formality. In studying solution concentration, Sinaumed’s can read Solution Chemistry books by Rusman, Ratu Fazlia Inda Rahmayani, and Mukhlis which can help you to develop knowledge about solution concentration.
Solution Concentration Formula
It turns out that the concentration of the solution itself has a formula, so you shouldn’t calculate it arbitrarily. That way, the substance to be dissolved will be maximized. The formula for the concentration of the solution is C = m/V. In this formula, C is the concentration, m is the mass of the solute, and V is the total volume of the solution.
Solution Concentration Unit
Below will be explained further about the concentration of the solution along with how to calculate the concentration of the solution.
1. Molarity ( M )
Molarity in the concentration of a solution is known as molar concentration or molarity with the symbol it has, namely M. Molarity is used to quantitatively obtain the concentration of a solution. Expressed as the number of moles of a solute in solution divided by the specified volume of solution in liters. The following is the molarity formula for the concentration of a solution.
|Molarity (M) = moles of soluteliters of solution
|Molarity ( M ) = mass of solute g Mr solute 1000volume of solution (mL)
Question: 2.00 grams of sodium hydroxide, NaOH, is dissolved in water to form a solution with a volume of 200 mL. What is the molarity of NaOH in solution?
In order to know the amount of molarity, you must look at the ratio between the number of moles of solute and the number of liters of solution. In other words, we know the amount of NaOH in moles and the volume in liters.
The formula mass of NaOH is 40.0 g/mol, thus:
|2.00 g NaOH 1 mol NaOH40.0 g NaOH = 0.500 mol NaOH
If expressed in liters, 200 mL becomes 0.200 L
|Molarity (M) = 0.05 mol NaOH0.2 L solution
= 0.250 mol NaOH/liter
= 0.250 M NaOH
So, the concentration of NaOH in the solution is 0.250 M NaOH.
Why is molarity so important in the concentration of a solution? Because, if we know the molarity of a solution, then we can determine the desired number of moles. To find out the desired number of moles, you must measure the volume precisely.
2. Molality (m)
Molality is a concentration of a solution that states the number of moles of a compound or substance per kilogram of solvent ( solvent ).
If, Mm is the molar mass (g mol-1) then:
|molality m=m (solute) Mm x kg of solvent
|molality m=mass of solute (g)Mr of solute x 1000mass of solvent (g)
Question: Calculate the molality of a methyl alcohol solution (Mr = 32) by dissolving 37 g of methyl alcohol (CH3OH) in 1750 g of water.
|moles of solute = 37 g32.0 gmol = 1.156 mol
After getting the mass of the solvent, then find the molality.
|molality = 1.156 mol1.750 kg=0.660 m
3. Normality (N)
Normality can be defined as the number of equivalent moles of a substance per liter of solution. So, the normality formula can be written as follows:
|normality = V molecule (liters)
Meanwhile, the normality formula used to find solids dissolved in water is different from the normality formula for the number of equivalent moles of a substance per liter. The following is the formula for normality of solids dissolved in water.
|normality=grams of solute equivalent mass x liters of solution
N= m (X)Mm Xx x 1000V (ml) x
Meanwhile, the equivalent relationship and molecular weight can be formulated as follows:
|gram equivalent= Mrn
Where n is the number of equivalents per mole of substance X.
Question: Calculate the normality of a solution containing 36.75 g, H2SO4 in 1.5 liters of solution. Mr H2SO4 = 98.
Equivalent mass = 49
|normality = 36.7549 x 1.50 = 0.50 n
4. Mole fraction
The mole fraction is used to express the moles of a substance per the total number of moles. The following is the mole fraction formula.
|mole fraction A = X A = N (A)N (total)
|mole fraction of solute = number of moles of solute, number of moles of solute + number of moles of solvent
|mole fraction of solvent = number of moles of solvent, number of moles of solute + number of moles of solvent
Question: If it is known that Mr. H2O = 18 g/mol. What is the mole fraction of H2SO4 in a solution containing 0.56 moles in 1 kg of H2O?
|number of moles of water = 100018
= 55.56 moles
|mole fraction of H 2 SO 4 = 0.56056+55.56
5. Concentration in percent
In chemistry, the term percent is often used to express the concentration of a solution. Percent in solution concentration can be expressed in three forms, namely weight percent (%W/W), volume percent (%V/V), and volume weight percent (%W/V). Weight percent is often used because it does not depend on temperature.
a, Percent by weight (%W/W)
|weight percent (%W/W)=grams of solutegrams of solutionx 100
Calculate how many % by weight of NaCl is made to dissolve 10 grams of NaCl in 50 grams of water.
|Weight % NaCl= 1010+50 x 100
b. Percent of volume (%V/V)
|volume percent (%V/V)= mL of solute mL of solution x 100
40 mL of alcohol is mixed with 40 mL of water to give 86.54 mL of solution. What is the volume percent of each component?
|percent by volume of alcohol %VV= 40 mL86.54 mLx 100
|percent by volume of water %VV=40 mL86.54 mLx 100
c. Percent by volume (%W/V)
|weight percent by volume (%W/V)= grams of solutemL solution x 100
Question: CH3COOH in 5 mL of vinegar at the rate of 1.008 g/mL containing 0.2589 g CH3COOH. What is the volume percent by weight?
|% wt volume = 0.2589 g5 mL x 1.008 gmL x 100
6. Parts Per Million (ppm) and Parts per Billion (ppb)
Both units of solution concentration are commonly used when the solution becomes dilute. The concentration of parts per million is parts per one million and parts per billion is parts per one billion. Both of these units are widely used to analyze trace elements with SSA (Atomic Absorption Spectrometry).
The ppm unit expresses the number of grams of a substance in 106 grams of solution. Check out the Parts Per Million (ppm) formula.
|ppm= m (substance)m (sample) x 106 ppm
|1 ppm= 1 mg of solute in 1 L of solution
The ppb unit expresses the number of grams of a substance in 109 grams of solution. Check out the Parts Per Billion (ppb) formula.
|ppb= m (substance)m (sample) x 109 ppb
|1 ppb= 1 μ g solute1 L solution
Question: There is an acetone in water solution which contains 8.60 mg of acetone in 21.4 liters of solution. If the solubility density is 0.997 g/cm3, then calculate the concentration of acetone in a (ppm) and b (ppb).
Weight of acetone 8.60 mg = 8.60 x 10-3 g
Weight of water = 21.4 L x 1000 mL/L x 0.997 g/mL
= 21.4 x 104 g
|ppm acetone = 8.60 g acetone21.4 x 104 g water x 106 ppm
= 0.402 ppm
|ppb acetone = 8.60 g acetone21.4 x 109 g water x 109ppb
= 402 ppb
7. Formal (F)
Formality is the ratio between the total mass of the solute formula in each liter of solution. Formality can be referred to as the actual concentration that comes from the solute or dissolved ions present in the solution. Because of this, there are differences as well as similarities between formality and formality. The following is the formal formula.
|formality F= total mass of solute formulaliters of solution
Question: A solution is prepared to dissolve 1.90 g of Na2SO4 and this solution is in 0.085 liters of solution. How much formality of the solution?
Molar mass, Na2SO4 : 142 g/mol
|1.90 g= 1.90 g142 gmol=0.0134 formula weight
After getting the total mass of the formula, then the formality is calculated.
|formality = 0.01340.085 = 0.16 F
Sinaumed’s can learn more about solution concentrations through the SMA/MA Class 1,2,3 book Tentor’s Chemistry Bimbel Method which is made by summarizing various difficult materials, so it will be easier to understand chemistry and in which there are examples of questions and discussion.
SOLUTION CONCENTRATION IN CHEMICAL AND PHYSICS UNITS
Basically, the units of concentration of solutions that have been mentioned are divided into two groups, namely physical units and chemical units. The following table of physical and chemical units.
|Concentration of Solutions in Chemical Units
|moles of solute liters of solution
|moles of solute kg of solvent
|solute equivalentliters of solution
|moles of solute liters of solution
|the formula mass of the solute in liters of solution
|One-thousandth of a mole of charge
|osmolsliter of solution
|Concentration of Solutions in Physics
|grams of solute grams of solution x 100
|mL of solute mL of solution x 100
|Volume weight percent
|grams of solute mL of solution x 100
|mg of solute 100 mL of solution x 100
|Parts per million
|1 mg of solute 1 L of solution
|Parts per billion
|1 μg of solute1 L of solution
How to Make a Solution
Procedure for preparing a solution in solid form. Check out the steps.
- The number of moles required must be calculated in advance so that the solution is in accordance with its volume and concentration.
- Determine the molar mass of the compound used to calculate the required mass.
- The next step is to place the watch glass on the scales. Then, set the scale at “0”. Weigh it carefully in order to get the required weight from the mass of the substance.
- After weighing the substance, transfer the substance into the beaker. Then, add water to dissolve the substance. So that no substance is left behind, clean the watch glass by rinsing it with clean water, then transfer the rinse water to a beaker. Do the rinse at least twice.
- Stir using a stir bar until all the substance is dissolved, then transfer the solution into a measuring flask. To clean up residue in the beaker and stir bar, it’s best to use water from a wash bottle.
- The next step is to add water to the volumetric flask up to the mark, then add the last drop of water with a dropper to ensure that the base of the meniscus is right at the boundary line.
- Close the volumetric flask and shake the volumetric flask several times so that the solution is mixed evenly.
- The final step is to label the solution with the name of the solution and the date of manufacture.
Primary and Secondary Standard Solutions
In chemistry, every solution has a standard. A standard solution is a solution whose concentration is known with certainty. If, using a standard solution, it will be able to determine the concentration of another solution. Standard solutions have uses that can be utilized when in a chemical laboratory. Following are some of the uses of standard solutions.
- Serves to standardize volumetric solutions
- Can be used as a reference to determine a solution whose concentration is unknown.
- To calibrate the instrument
- Prepare standard solution.
Standard solutions are divided into two types, namely primary standard solutions and secondary standard solutions.
1. Primary Standard Solution
Primary standard solution is a solution whose concentration is obtained from the measurement results. The primary standard is used as the primary calibrator.
The requirements for a primary standard solution are to be stable, anhydrous, not hygroscopic, have high purity, have a high molecular level, when compared to similar molecules, are inexpensive, are not toxic, and are always available and ready to use.
2. Secondary Standard Solution
Secondary standard solution is a solution whose concentration is obtained by determining the concentration of a substance using a primary standard solution. Secondary standard solutions are commonly used for calibration purposes of control materials in small laboratories to analyze unknown concentrations of solutions of a substance.
The requirements for a secondary standard solution, namely the level of purity of the solution is lower when compared to the primary standard solution, the solution has a stable time, determines the level of a substance using a primary standard solution, and has lower stability when compared to a primary standard solution.
Solution in chemistry has the meaning of a homogeneous mixture with a composition ratio according to its constituent components. In chemistry, solutions are divided into two types of standards, namely primary standards and secondary standards. The concentration of a solution is the amount of substance dissolved in each unit of solution or solvent.
In simple terms, the concentration of a solution can provide an overview or information about the ratio of the amount of solute to the amount of solvent. The concentrations of solutions commonly used in laboratories are Molarity, Molality, Normality, Mole Fraction, Concentration in Percent, Parts per Million (ppm) and Parts per Billion (ppb), and Formality.
So, that’s an explanation of the concentration of a solution, starting from the definition, types, and units that Sinaumed’s needs to know. Sinaumed’s will often find material on the concentration of solutions in chemistry and physics lessons from elementary school. So Sinaumed’s can understand this material better so that it is easier to find out about a wider range of material solutions and can more easily solve questions.