Biochemistry Laboratory
Solutions are homogeneous mixtures of two or more substances. Components of a solution in general; dissolving matter and dissolved substance or substances. The solvent is usually liquid and water; It can also be liquids such as alcohol, chloroform. Dissolved substances can be solid, liquid, gas.
Concentration relates to the components of a mixture or solution. Concentration refers to the amount of substance in a given area. In other words, concentration is the ratio of the solute in a solution to the solvent or the total solution. Concentration is expressed in mass per unit volume. The concentration of the solute can also be expressed in units of moles or volume. The concentration of chemical solutions can be calculated for any mixture.
How to Calculate Concentration
Concentration is determined mathematically by taking the mass, moles, or volume of the solute and dividing it by mass, mole, or solution volume. Some units of concentration and examples include:
1) Mass percent (%w/w): Mass percent is the mass of the element or solute divided by the mass of the compound or solute. The result is multiplied by 100 to give a percent.
Mass Percent= Mass of Solute /Mass of Solution ×100%
Example 1.1: A salt solution with a mass of 365 g has 36.5 g of NaCl dissolved in it. What is the mass / mass percent concentration of the solution?
mass percent = (grams solute / grams solution) x 100
% m / m = 36.5 g / 365 g × 100% = 10%
Example 1.2: 8 g NaOH dissolved in 80 g of water. What is the mass / mass percent concentration of the solution?
mass percent = (grams solute / grams solution) x 100
grams solution = 8 g sodium hydroxide + 80 g water=88
%m/m = 8/ 88g × 100% = 11%
Example 1.3: what is masses of sodium chloride (NaCl) and water required to obtain 175 g of a 15% solution?
mass percent = (grams solute / grams solution) x 100
15% = (x grams sodium chloride / 175 g total) x 100
x = 15 x 175 / 100
x = 26.25 grams NaCl
mass of water = total mass - mass of salt
mass of water = 175 g - 26.25 g
mass of water = 147.75 g
2) Percent weight by volume (%w/v): Percent of weight of solution in the total volumeof solution. Percent here is the number of grams of solute 100 mL of solution.
Example 2.1: What is the weight/volume percentage concentration of 150 mL of aqueous sodium chloride solution containing 20g NaCl?
weight/volume (%) = (mass solute / volume of solution) × 100
w/v (%) = (20g / 150 mL) × 100 = 7.5g/100mL (%)
Example2.2: How many grams of NaOH would be needed to make a 40% w/v solution using 100 mL water as the solvent?
weight/volume (%) = (mass solute / volume of solution) × 100
40% = (x grams NaOH / 100ml) x 100
x = 40x 100 / 100
x = 40 grams NaOH
Example 2.3: 10 g BaCl2 is dissolved in 90 g of water. The density of the solution is 1.09 g/mL. What is the weight/volume percentage concentration of the solution?
weight/volume (%) = (mass solute / volume of solution) × 100
density = mass(solution) / volume(solution)
mass(solution) = mass(solute) + mass(solvent)
mass(solution) = 10 g BaCl2 + 90 g water = 100.00 g
volume solution = mass / density
volume solution = 100 g / 1.09 g/mL = 91.74 mL
w/v (%) = (10.00 g ÷ 91.74 mL) × 100
w/v (%) = 10.90 g/100 mL (%)
3) Hacimce yüzde hacim (% v / v): The ratio of the volume of solute to the total volume of the solution is expressed by multiplying by 100.
Example 3.1: What is the % v/v of a solution that has 5.0 mL of hydrochloric acid (HCl) diluted to 100 mL with water?
Percent by volume = volume of solute / total volume of solution × 100 %
% v/v = 5 mL HCl/100 mL of solution= 5%
Example 3.2: : what is volume of sodium chloride (NaCl) and water required to obtain 225 mL of a 20% solution?
Percent by volume = volume of solute / total volume of solution × 100 %
20%= x volume of solute / 225 × 100 %
X=20×225 / 100 = 50 %
Example 3.3: 2 L of an aqueous solution of HCl contains 450 mL of HCl. What is the volume/volume percentage concentration of this solution in mL/100mL?
Percent by volume = volume of solute / total volume of solution × 100 %
volume of solution = 2 L = 2 L × 1000 mL/L = 2000 mL
v/v (%) = [450 / 2000 mL] × 100 = 22.5 mL/100mL (%)
4) Percent volume by weight (%v/w): Percent of volume of solution in the total of weight of solution.
Example 4.1: What is the weight/volume percentage concentration of 250 mL of aqueous sodium chloride solution containing 5 g NaCl?
weight/volume (%) = (mass solute / volume of solution) × 100
w/v (%) = (5g / 250 mL) × 100 = 2g/100mL (%)
5) Molarity: Molarity is the number of moles of substance dissolved in a liter of solution. Its unit is molar. It is indicated by the M symbol.
Example 5.1: What is the molarity of 245 g of H2SO4 dissolved in 1.000 L of solution?
M = moles of solute / liters of solution
MV = grams / molar mass
(x) (1.000 L) = 245 g / 98. g mol¯1
x = 2.5 M
Example 5.2: What volume in mL of 12.0 M HCl is needed to contain 3.00 moles of HCl?
M = moles of solute / liters of solution
MV = grams / molar mass
12.0 M = 3.00 mol / x
x = 0.250 L
0.250 L x (1000 mL / L) = 250. mL
Example 5.3: A solution of calcium bromide contains 20.0 g dm-3. What is the molarity of the solution with respect to calcium bromide and bromine ions.
MV = mass / molar mass
(x) (1.00 L) = 20.0 g / 199.886 g/mol
x = 0.100 M
When CaBr2 ionizes, two bromide ions are released for every one CaBr2 that dissolves. That leads to this:
[Br-] = 0.200 M
6) Normality: Normality is a measure of concentration equal to the gram equivalent weight per liter of solution. Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality.
Example 6.1: What is the normality of the 0.1381 M NaOH ?
N = Molarity (M) × number of equivalents(n)
N = 0.1381 mol/L × (1 eq/1mol) = 0.1381 eq/L = 0.1381 N
Example 6.2: Calculate the normality of 0.321 g sodium carbonate when it is mixed in a 250 mL solution.
N = Molarity (M) × number of equivalents(n)
N of 0.321 g sodium carbonate
N = Na2CO3 × (1 mol/105.99 g) × (2 eq/1 mol)
N = 0.1886 eq/0.2500 L
N = 0.0755 N
Example 6.3: 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?
N = Molarity (M) × number of equivalents(n)
Initial Normality (N1) × Initial Volume (V1) = Normality of the Final Solution (N2) × Final Volume (V2)
Na × (25,00 mL) = (0,1718N) (28,12 mL)
Therefore, the concentration of citric acid= 0.1932 N.
7) Molality: Molality is the number of moles of the substance dissolved in one kilogram of solvent. The unit is break. It is indicated by the symbol m. It is expressed in Mol / Kilogram.
Example 7.1: 7.45 potassium chloride (KCl) was dissolved in 100g of water. Colculate the molality of the solution.( molecular mass of KCl=74.5 g /mol)
Number of moles of KCl= 7.45 / 74.5 g/mol = 0.1 mol
Molality(m): moles of solute / kilograms of solvent
M= 0.1 mol /0.1 kg = 1 mol /kg
Example 7.1: A solution of H2SO4 with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?
Molality(m): moles of solute / kilograms of solvent
8.010 m means 8.010 mol / 1 kg of solvent
8.010 mol times 98.0768 g/mol = 785.6 g of solute
785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution.
1785.6 g divided by 1.354 g/mL = 1318.76 mL
8.01 moles / 1.31876 L = 6.0739 M
Example 7.3: Calculate the molality of 15.00 M HCl with a density of 1.0745 g/cm3
15.00 mol/L times 1.000 L = 15.00 mole of HCl
15.00 mol times 36.4609 g/mol = 546.9135 g of HCl
1000. mL times 1.0745 g/cm3 = 1074.5 g of solution
1074.5 g minus 546.9135 g = 527.5865 g of water = 0.53 kg
15.00 mol / 0.5275865 kg = 28.43 m
8) Mole Fraction: Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. It’s a way of expressing the concentration of a solution.
Example 8.1: what is the mole fraction of CH3OH and H2O in a solution prepared by dissolving 5.5 g of alcohol in 40 g of H2O. M of H2O is 18 and M of CH3OH is 32.
Moles of CH3OH = 5.5 / 32 = 0.17 mole
Moles of H2O = 40 / 18 = 2.2 moles
Therefore, according to the equation
mole fraction of CH3OH = 0.17 / 2.2 + 0.17
mole fraction of CH3OH = 0.073
Example 8.2: 23 g of ethyl alcohol is dissolved in 54 g of water . colculate the mole fraction of ethyl alcohol. (molar mass; ethyl alcohol:46 g/mol , water: 18 g/mol)
Mole fraction of solute= Number of moles of solute / number of moles of solute + number of moles of solvent
Number of moles ethyl alcohol = 23g / 46 g/mol=0.5 mol
Number of moles water= 54g / 18g /mol=3 mol
Mole fraction of solute =0.5/3.5= 0.871
Example 8.3: what is mole fraction of HCl in a solution of HCl containing 24.8% of HCl by mass. (molecular mass of HCl:3605g)
Percentage by mass of HCl =24.8%
Let us consider 100g of HCl solution
Mass of HCl 24.8 g
Mass of solvent = 100 – 24.8= 75.2g
Moles of HCl =24.8g/ 36.5g = 0.6795 mol
Number of moles water= 75.2g / 18g =4.178 mol
Total number of moles = 4.178+ 0.6795= 4.8575
Mole fraction of HCl = 0.6795 / 4.8575= 0.1399
9) Parts per million: His is an abbreviation for "parts per million" and it also can be expressed as milligrams per liter (mg/L). This measurement is the mass of a chemical or contaminate per unit volume of water.
Example 9.1: A solution has a concentration of 0.033 g kg-1. What is its concentration in ppm?
ppm = mass solute (mg) / volume solution (L)
Convert the mass in grams to mass in milligrams:
0.033 g = 0.033 g × 1000 mg / g = 33 mg
Rewrite the concentration in mg / kg
concentration in mg kg-1 = 33 mg/ kg = 33 ppm
10) Parts per billion: Parts per billion (ppb) is the number of units of mass of a contaminant per 1000 million units of total mass. Also µg/L or micrograms per liter.
Example 10.1: 0.025 gram of Pb(NO3)2 is dissolved in 100. grams of H2O, what is the concentration of the resulting solution, in parts per million?
ppb = mass solute (mg) / volume solution (L)
0.025g / (100g + 0.025) x 1,000,000= 250 ppm
11) Parts per trillion: Parts per trillion (ppt) is the number of units of mass of a contaminant per 10000000 million units of total mass.
1 ppm = 1000 ppb = 1000000 ppt
Example 11.1: 500 mL of 600 ppt solution of sucrose. What volume of this solution in millilitres contains 0.15 g of sucrose?
ppt = mass solute (mg) / volume solution (L)
volume solution (L) = mass solute (mg) /ppt
mass solute (sucrose) required = 0.15 g
ppt concentration of solution = 600 ppt = 600 mg / L
mass solute sucrose = 0.15 g × 1000000000 mg/g = 150000000 mg
volume solution (L) required = 150000000 mg / 600 mg /L = 250000 L
volume solution required = 250000 L × 1000 mL/L = 250000000 mL
