Volumetric Analysis: Acid-Base Chpt. 13
Quantitative Analysis: is analysis which involves investigating the quantities or amounts of materials present. Gravimetric analysis (Chpt. 11) – composition of substances determined by careful weighing Volumetric analysis (Chpt. 13) – composition of substances determined by reacting together volumes of solutions
REMEMBER: A solution is a mixture of a solute and a solvent A solvent: is a substance that dissolves other materials A solute: is the substance that dissolves in the solvent
Concentration: A concentrated solution contains a large amount of solute per litre of solution e.g. strong coffee A dilute solution contains a small amount of solute per litre of solution e.g. weak coffee
Concentration The concentration of a solution is the amount of solute that is dissolved in a given volume of solution
There are several ways of expressing the concentration of a solution: • Percentage of solute – 3 forms • Parts per million (ppm) • Moles of solute per litre of solution (MOLARITY)
Percentage of solute • (This method of expressing concentration is usually used in many household solutions and in medicine) • There are 3 ways in which the percentage of solute in a solution is expressed: • a) Percentage weight per weight (w/w) • b) Percentage weight per volume (w/v) • c) Percentage volume per volume (v/v)
Percentage weight per weight (w/w): • This is the number of grams of solute per 100g of solution e.g. • 10% w/w NaCl → 10g of sodium chloride per 100g of solution • 2% w/w Arnica ointment → 2g of arnica per 100g of ointment
Percentage weight per volume (w/v): • This is the number of grams of solute per 100cm3 of solution e.g. • 10% w/v NaCl → 10g of sodium chloride per 100cm3 of solution • 5% w/v NaCl → 5g of NaCl in 100cm3 of solution • *Note: usually the units are grams per litre (g/L) therefore you would have to bring it to a litre*
Percentage volume per volume (v/v): • This is the number of cm3 of solute per 100cm3 of solution e.g. • 5% v/v vinegar solution → 5cm3 of ethanoic acid per 100cm3 of vinegar • 13% v/v wine solution → 13cm3 of ethanol (alcohol) per 100cm3 of wine
Complete the following table: *Note: DO NOT take down table
The following calculations involve working with percentages!!!! Make sure you understand the definitions!!!!!
Example 1: A solution contains 20g of potassium hydroxide in 1 litre of solution. Express the concentration of the solution in % (w/v). Solution:
Example 2: A bottle of vinegar contains 25cm3 ethanoic acid in 500cm3 of solution. Express the concentration of ethanoic acid in the solution in % (v/v) Solution:
Example 3: A solution contains 10g of sodium carbonate in 40g of solution. Express the concentration of the solution in % (w/w). Solution:
Example 4: The label on a bottle of wine indicates that the concentration of alcohol in the wine is 9% (v/v). What volume of alcohol is there in 250cm3 of the wine? Solution:
Try the following: • A sample of sea water has a mass of 1.3kg. On evaporating the water, 148g of salt was recovered from it. Express the concentration of the salt as % w/w. • Some illnesses can upset the salt balance in the body and it may be necessary to administer salt intravenously. The solution of salt that is injected is marked 0.85% w/v. What weight of sodium chloride is needed to make up 250cm3 of this solution?
2. Parts per million (ppm) • This method of expressing the concentration of a solution is only used for very dilute solutions i.e. when dealing with very low concentrations of substances. • This is the number of milligrams per litre (mgL-1) • *Note: 1 Litre of water has a mass of 1 million milligrams* • So, can say 1mg/L = 1 mg per million mg • = 1 ppm • Example: the concentration of chlorine in water is • 2 ppm this means there are 2 mg of chlorine per • litre of water
Example 1: 1 gram = 1000 milligrams
3) Moles of solute per litre of solution (MOLARITY) Remember: One mole of a substance is the amount of that substance which contains 6 x 1023 particles (atoms, ions, molecules) of that substance Mass of 1 mole of an element = Relative Atomic Mass in grams e.g. 1 mole of Na = 23 g 1 mole of Mg = 24g
The most important way of expressing the concentration of a solution is in terms of moles per litre of solution (molarity) • Definitions: • The Molarity of a solution is the number of moles • of solute per litre of solution • A 1 molar solution is a solution which contains one • mole of solute per litre of solution • also, • - a solution which contains 2 moles of • solute in a litre of solution is said to be 2 • molar (2M) • - a solution which contains 0.5 moles of solute in a litre of solution is said to be 0.5 molar (0.5M)
Symbols Used: • - M • - mol/L or mol L-1 Remember: No. of Moles of = Mass of Substance Substance Molar Mass
Concentration Examples: -1 mol/L NaOH = 40g NaOH(MrNaOH = 40) per litre of solution - 2mol/L NaOH = 80g NaOH per litre of solution - 0.5mol/L NaOH = 20g NaOH per litre of solution - 0.1 M (decimolar) NaOH = 4g NaOH per litre of solution
Complete the following: 1M H2SO4 = 0.5M H2SO4 = 3M H2SO4 =
Calculations Involving Molarity Three types: 1. Converting Molarity to Grams per Litre 2. Converting Grams per Litre(Volume) to Molarity 3. Calculation of number of Moles from Molarity and Volume
Converting Molarity to Grams per Litre • Example 1: • What is the concentration in g/L of a 0.1 M H2SO4 • Solution? • Example 2: • How many grams of NaCl per litre are present in a solution • marked 0.25 M NaCl. • Example 3: • Calculate the concentration in grams per litre of • bench dilute sulphuric acid whose concentration Is 1.5mol/L Concentration in g/L = Molar Mass x Molarity
2. Converting Grams per Litre(Volume) to Molarity Molarity = Grams per Litre Molar Mass Example 1: What is the molarity of a NaOH solution containing 4g of NaOH per litre? Solution:
Example 2: What is the molarity of a solution that contains 3.68g of NaOH per litre of solution?
Example 3: Calculate the concentration in moles per litre of a solution containing 45 grams of sulphuric acid per 250cm3 of solution.
3. Calculation of number of moles from molarity and volume No. of moles = Volume(L) x Molarity Example 1: How many moles are there in 250cm3 of 0.1 M HCl? Solution:
Example 2: How many moles of NaOH are present in 25cm3 of 0.55M NaOH
Example 3: How many moles of hydrochloric acid are present in 30cm3 of 0.2M HCl
Example 4: What mass of sodium hydroxide is contained in 25cm3 of a 0.75M solution of sodium hydroxide?
Example 5: What volume of 0.15M sodium hydroxide solution will contain 5 grams of sodium hydroxide?
Balanced Chemical Equations A balanced equation tells you the amounts of substances that react together and the amounts of products formed. Consider the following balanced equation for the reaction between hydrogen gas and oxygen gas to form water: 2H2 + O2 → 2H2O This equation can be interpreted in a number of ways.
In terms of molecules: 2 molecules 1 molecules 1 molecule In terms of Avogadro’s number of molecules: 2 x 6 x 1023 1 x 6 x 1023 2 x 6 x 1023 molecules moleculesmolecules *REMEMBER: the amount of a substance which contains the Avogadro number of particles is called a mole of that substance 2H2 + O2 → 2H2O 2H2 + O2 → 2H2O
In terms of Moles of a substance: 2 moles 1 mole 2 moles Further examples: a) 2Mg + O2 → 2MgO 2 moles 1 mole 2 moles b) CaCO3 → CaO + CO2 1 mole 1 mole 1 mole c) CH4 + 2O2 → CO2 + 2H2O 1 mole 2 moles 1 mole 2 moles 2H2 + O2 → 2H2O
Reactions between a solution and a solid In a number of chemical reactions solids react with solutions. You may be asked to calculate the mass of metal which reacts with a given volume of acid Example 1: What mass of magnesium will react with 50cm3 of 0.5M H2SO4. The balanced equation for the reaction is: Mg + H2SO4 → MgSO4 + H2
Example 2: Sodium carbonate, Na2 CO3, reacts with dilute hydrochloric acid according to the equation: Na2CO3 + 2HCl → 2NaCl + CO2 + H2O What volume of hydrochloric acid of concentration 0.75M would be needed to neutralise 7.5g of anhydrous sodium carbonate?
Concentration of Solutions 1 mole/250cm3 1 mole/L 1 mole/500cm3 1 mole/100cm3
If in each volumetric flask one mole of solute is dissolved then as the volume becomes smaller, the concentration increases. In the case of a coloured solution , as the concentration increases, the intensity of the colour also increases (see diagram pg. 148 book)
Dilution of Solutions To save space in our prep room we buy solutions in concentrated form, i.e. 18M HCl (18 mol L-1). We call these stock solutions Definition: The process of adding more solvent to a solution is called dilution. A typical dilution involves determining how much water must be added to an amount of stock solution to achieve a solution of the desired concentration
When a solution is diluted, more solvent is added but the quantity of solute is unchanged: Moles of solute = Moles of solute before dilution after dilution Since the volume of the solution increases and the number of moles present remains the same, then the concentration of the solution must decrease *Note: diluting a coloured solution results in a lightening of the colour of the solution i.e. colour intensity is proportional to concentration
Calculation of the Effect of Dilution on Concentration MolarityDil x VolumeDil= MolarityConc.x VolumeConc. M1 x V1 = M2 x V2
Example 1: If 20cm3 of a 3M hydrochloric acid solution is diluted to a volume of 1 L with water, what is the concentration of the diluted acid? Solution:
Example 2: What volume of a 2M sodium hydroxide solution is needed to make up 100 cm3 of a 0.1 M sodium hydroxide solution
Student Questions:Question 1: If 12cm3 of a 0.1M sodium hydroxide solution is diluted to a volume of 500cm3 with water, what is the concentration of the diluted solution? Question 2: What volume of 1M NaOH solution is needed to make 300cm3 of 0.05M solution?
Standard Solutions Definition: A standard solution is a solution whose concentration is accurately known e.g. a solution containing 10 grams of NaCl per litre is a standard solution In the determination of the concentration of an acid a standard solution of an alkali is used and to determine the concentration of an alkali a standard acid would be used. However, before any determinations can be made a starting accurately standardised solution is required – from which to find the exact concentration of other solutions