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Volumetric Analysis: Acid-Base Chpt. 13

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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

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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

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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

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Concentration The concentration of a solution is the amount of solute that is dissolved in a given volume of solution

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There are several ways of expressing the concentration of a solution: 1)Percentage of solute – 3 forms 2)Parts per million (ppm) 3)Moles of solute per litre of solution (MOLARITY)

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1.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)

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a)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

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b)Percentage weight per volume (w/v): This is the number of grams of solute per 100cm 3 of solution e.g. 10% w/v NaCl 10g of sodium chloride per 100cm 3 of solution 5% w/v NaCl 5g of NaCl in 100cm 3 of solution *Note: usually the units are grams per litre (g/L) therefore you would have to bring it to a litre*

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c)Percentage volume per volume (v/v): This is the number of cm 3 of solute per 100cm 3 of solution e.g. 5% v/v vinegar solution 5cm 3 of ethanoic acid per 100cm 3 of vinegar 13% v/v wine solution 13cm 3 of ethanol (alcohol) per 100cm 3 of wine

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Complete the following table: Concentration Unit ExampleMeaning Percentage weight per volume (w/v) 3% (w/v) NaCl solution Percentage volume per volume (v/v) 3% (v/v) alcohol solution Percentage weight per weight (w/w) 3% (w/w) sugar solution *Note: DO NOT take down table

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The following calculations involve working with percentages!!!! Make sure you understand the definitions!!!!!

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Example 1: A solution contains 20g of potassium hydroxide in 1 litre of solution. Express the concentration of the solution in % (w/v). Solution:

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Example 2: A bottle of vinegar contains 25cm 3 ethanoic acid in 500cm 3 of solution. Express the concentration of ethanoic acid in the solution in % (v/v) Solution:

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Example 3: A solution contains 10g of sodium carbonate in 40g of solution. Express the concentration of the solution in % (w/w). Solution:

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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 250cm 3 of the wine? Solution:

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Try the following: 1.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. 2.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 250cm 3 of this solution?

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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

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Example 1: 1 gram = 1000 milligrams

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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 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

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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)

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Symbols Used: - M - mol/L or mol L -1 Remember: No. of Moles of= Mass of Substance Substance Molar Mass

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Concentration Examples: -1 mol/L NaOH = 40g NaOH (M r NaOH = 40) per litre of solution - 2mol/L NaOH = 80g NaOH per litre of solution - 0.5mol/L NaOH = 20g NaOH per litre of solution M (decimolar) NaOH = 4g NaOH per litre of solution

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Complete the following: 1M H 2 SO 4 = 0.5M H 2 SO 4 = 3M H 2 SO 4 =

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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

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1.Converting Molarity to Grams per Litre Example 1: What is the concentration in g/L of a 0.1 M H 2 SO 4 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

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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:

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Example 2: What is the molarity of a solution that contains 3.68g of NaOH per litre of solution?

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Example 3: Calculate the concentration in moles per litre of a solution containing 45 grams of sulphuric acid per 250cm 3 of solution.

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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 250cm 3 of 0.1 M HCl? Solution:

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Example 2: How many moles of NaOH are present in 25cm 3 of 0.55M NaOH

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Example 3: How many moles of hydrochloric acid are present in 30cm 3 of 0.2M HCl

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Example 4: What mass of sodium hydroxide is contained in 25cm 3 of a 0.75M solution of sodium hydroxide?

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Example 5: What volume of 0.15M sodium hydroxide solution will contain 5 grams of sodium hydroxide?

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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: 2H 2 + O 2 2H 2 O This equation can be interpreted in a number of ways.

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In terms of molecules: 2 molecules1 molecules 1 molecule In terms of Avogadros number of molecules: 2 x 6 x x 6 x x 6 x molecules molecules molecules *REMEMBER: the amount of a substance which contains the Avogadro number of particles is called a mole of that substance 2H 2 + O 2 2H 2 O

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In terms of Moles of a substance: 2 moles1 mole2 moles Further examples: a)2Mg + O 2 2MgO 2 moles 1 mole 2 moles b)CaCO 3 CaO + CO 2 1 mole 1 mole 1 mole c) CH 4 + 2O 2 CO 2 + 2H 2 O 1 mole 2 moles 1 mole 2 moles 2H 2 + O 2 2H 2 O

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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 50cm 3 of 0.5M H 2 SO 4. The balanced equation for the reaction is: Mg+ H 2 SO 4 MgSO 4 + H 2

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Example 1 Solution:

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Example 2: Sodium carbonate, Na 2 CO 3, reacts with dilute hydrochloric acid according to the equation: Na 2 CO 3 + 2HCl 2NaCl + CO 2 + H 2 O What volume of hydrochloric acid of concentration 0.75M would be needed to neutralise 7.5g of anhydrous sodium carbonate?

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Example 2 Solution:

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Concentration of Solutions 1 mole/L 1 mole/500cm 3 1 mole/250cm 3 1 mole/100cm 3

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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)

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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

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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

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Calculation of the Effect of Dilution on Concentration Molarity Dil x Volume Dil = Molarity Conc. x Volume Conc. M 1 x V 1 = M 2 x V 2

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Example 1: If 20cm 3 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:

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Example 2: What volume of a 2M sodium hydroxide solution is needed to make up 100 cm 3 of a 0.1 M sodium hydroxide solution

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Student Questions: Question 1: If 12cm 3 of a 0.1M sodium hydroxide solution is diluted to a volume of 500cm 3 with water, what is the concentration of the diluted solution? Question 2: What volume of 1M NaOH solution is needed to make 300cm 3 of 0.05M solution?

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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

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A standard solution is prepared by weighing out a sample of solute, transferring it completely to a volumetric flask, and adding enough solvent (usually deionised water) to bring the volume up to the mark on the neck of the flask.

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Due to the fact that many substances can not be obtained in a high degree of purity standard solutions of common laboratory acids and bases cannot be prepared directly e.g.: - cannot weigh out 1 mole of sulphuric acid as it absorbs moisture from the air - cannot weigh out 1 mole of nitric acid as it is volatile - cannot weigh out 1 mole of iodine as it sublimes at room temperature In order to make up standard solutions substances which can be obtained in a highly pure state and which are stable in air are required

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Primary Standard Definition: A primary standard is a substance which can be obtained in a stable, pure and soluble solid form so that it can be weighed out and dissolved in water to give a solution of accurately known concentration Primary Standard Solution = Pure 100% Soluble Stable once made up Examples of Primary Standards: - Anhydrous sodium carbonate Na 2 CO 3 - Sodium Chloride NaCl - Potassium Dichromate K 2 Cr 2 O 7

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Mandatory Expt. 13.1: To prepare a standard solution of sodium carbonate Note: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each step

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Volumetric Analysis - Titrations Definitions: Standardise: means to find the concentration of a solution using titration A Titration: is a laboratory procedure where a measured volume of one solution is added to a known volume of another solution until the reaction is complete. Equivalence Point (End Point): the stage when the two solutions just react completely with each other

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Theory regarding apparatus and method involved in carrying out a titration on handout

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Mandatory Expt. 13.2: To use a standard solution of sodium carbonate to standardise a given hydrochloric acid solution Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each step Note 2: You must be able to carry out calculations on your results – see notes to follow *Need to calculate concentration of HCl in mol\L & g/L

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Volumetric Analysis Calculations 1.Calculating the unknown concentration of a solution from titration data 2.Calculating the relative molecular mass and the amount of water of crystallisation in a compound from titration data.

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1. Calculating the unknown concentration of a solution from titration data In straightforward titration calculations, where only the unknown concentration is required, the following formula can be used: V A x M A = V B x M B n A n B V A = volume (cm 3 ) of acid used V B = volume (cm 3 ) of base used M A = concentration of acid M B = concentration of base n A = no. of moles of acid in n B = no. of moles of base in balanced eqn for rxn balanced eqn for rxn

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Example 1: In a titration, 25cm 3 of a 0.05M sodium carbonate solution required 22cm 3 of a hydrochloric acid solution for complete neutralisation. Calculate the concentration of the hydrochloric acid solution. The equation for the reaction is: 2HCl (aq) + Na 2 CO 3(aq) 2NaCl (aq) + H 2 O (l) + CO 2(g)

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Example 1 Solution:

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Example 2: Hydrochloric acid and sodium hydroxide react according to the equation: HCl + NaOHNaCl + H 2 O 25cm 3 of a sodium hydroxide solution was titrated against a 0.2M HCl solution. The average titration figure was 23.5cm 3. Calculate the concentration of the sodium hydroxide solution in: a) mol/L b) g/L

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Example 2 Solution:

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Using results from your experiment calculate the concentration of the given hydrochloric acid solution in mol/L and g/L *Note: The first titration you performed was a rough titration which gave you an idea of where the end point is and so this result should be neglected. The remaining two titration results should agree within 0.1cm 3 of each other. The average of these results should be used in your calculation of the concentration of HCl.

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Example 3: 1.45g of sodium carbonate was dissolved in water and the solution was made up to 250cm 3 in a volumetric flask. 25cm 3 of this solution were titrated against a solution of hydrochloric acid using methyl orange as indicator. One rough and two accurate titrations were performed. The titration results and the equation for the reaction are given below. What is the concentration of the HCl solution in a) mol/L b) g/L Titration123 Volume Acid (cm 3) HCl (aq) + Na 2 CO 3(aq) 2NaCl (aq) + H 2 O (l) + CO 2(g)

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Example 3 Solution:

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Example 4: 10.0g of impure sodium hydroxide were weighed out, dissolved in water and solution made up to 250cm 3 in a volumetric flask. 25cm 3 of this solution, on being titrated with 1.1M HCl, required 21.8cm 3 of the acid for neutralisation. Calculate the % purity of the original sodium hydroxide.

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Example 4 Solution:

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Definition Secondary Standard: Make up a solution and then standardise this solution using a primary standard. This secondary standard can then be used to standardise other solutions e.g. HCl standardised and then used to standardise NaOH

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Mandatory Expt (Ordinary Level): A Hydrochloric Acid/Sodium Hydroxide titration and the use of this titration in making the salt sodium chloride Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each step Note 2: You must be able to carry out calculations on your results *Calculate concentration of NaOH in mol/L and g/L

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More Difficult Problems In a calculation involving a standard solution where more than just the unknown concentration is required, or where a solid is one of the reactants an alternative method from first principles should be used. Example 1: In a titration, 25cm 3 of a 0.12M NaOH solution required 24cm 3 of a H 2 SO 4 solution for complete neutralisation. Calculate: i) the number of moles of NaOH consumed ii) the number of moles of H 2 SO 4 consumed iii) the concentration of the H 2 SO 4 H 2 SO 4(aq) + 2NaOH (aq) Na 2 SO 4(aq) + 2H 2 0 (l)

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Example 1 Solution:

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Example 2: What mass of magnesium will react with 20cm 3 of a 0.09M hydrochloric acid solution? The equation for the reaction is: 2HCl (aq) + Mg (s) MgCl 2(aq) + H 2(g)

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Example 2 Solution:

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Applications of Acid-Base Titrations Mand. Expt.13.4: To determine the percentage of ethanoic acid in vinegar Mand. Expt. 13.5: To determine the percentage of water of crystallisation in hydrated sodium carbonate (washing soda)

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Mandatory Expt. 13.4: To determine the percentage of ethanoic acid (acetic acid) in vinegar Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each step Note 3: You must understand the need for and the use of the dilution factor Note 2: You must be able to carry out calculations on your results *Calculate concentration of ethanoic acid in the original vinegar in mol/L, g/L and % w/v

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Example 1: A sample of vinegar was diluted from 25cm 3 to 250cm 3 with water. In a titration, 25cm 3 of a 0.1M NaOH solution required 30cm 3 of the diluted vinegar for complete neutralisation. Calculate the concentration of ethanoic acid (CH 3 COOH) in the vinegar in: i) mol L -1 ii) g/L iii) %w/v The equation for the reaction is CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l)

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Example 1 Solution:

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2. Calculating the amount of water of crystallisation in a compound and the relative molecular mass of the compound from the titration data Definition: Water of crystallisation: is water that is chemically bound in the compound, which gives rise to the crystalline form. Crystals that contain water of crystallisation are said to be hydrated.

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Mandatory Expt. 13.5: To determine the percentage of water of crystallisation in hydrated sodium carbonate (washing soda) Hydrated sodium carbonate has the formula Na 2 CO 3.xH 2 O ( where x = no. of molecules of water of crystallisation present) The purpose of this experiment is to determine a value for x

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Note 1: You must have a clear understanding of all the steps you undertake in this experiment and be able to explain the importance of each step Note 2: You must be able to carry out calculations on your results *Calculate the molar mass of hydrated sodium carbonate, the percentage of water of crystallisation and the value of X in the formula Na 2 CO 3.xH 2 O Mandatory Expt. 13.5: To determine the percentage of water of crystallisation in hydrated sodium carbonate (washing soda)

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Mandatory Expt. 13.5: NOTE: Washing soda is composed of large translucent crystals of Na 2 CO 3.10H 2 O (form of hydrated sodium carbonate) It is important to realise that hydrated sodium carbonate gradually looses water of crystallisation over a period of time. Thus, if washing soda is used a lower value than the expected value of 10 for x in Na 2 CO 3.xH 2 O will be obtained.

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Example 1: Crystals of hydrated sodium carbonate ( Na 2 CO 3.xH 2 O) of mass 3.15g were dissolved in water and made up to 250cm 3 in a volumetric flask. 25cm 3 of this solution required 15cm 3 of a 0.15M HCl solution for complete neutralisation. The equation for the reaction is: 2HCl (aq) + Na 2 CO 3(aq) 2NaCl (aq) + H 2 O (l) + CO 2(g) Find the: i) concentration of the sodium carbonate solution ii) the molar mass of the sodium carbonate solution iii) the value of x ion the formula iv) the % of water of crystallisation in the hydrated sodium carbonate Na 2 CO 3.xH 2 O

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Example 1 Solution:

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Student Questions: Book: pg. 174 Nos & 3.16

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