ACID-BASE CALCULATIONS (LIVE) 02 JUNE 2015 Section A: Summary Notes

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1 ACID-BASE CALCULATIONS (LIVE) 02 JUNE 2015 Section A: Summary Notes Acids and Bases - Calculations Neutralisation Reaction Neutralisation is the reaction between an acid and a base. The end point is reached when the acid and base are chemically equivalent. An indicator is used during neutralisation reactions to show the end point. An indicator is an organic compound which turns a specific colour in an acid or base and at a specific ph. Titration is the name given to process of performing a neutralisation reaction. Concentration The unit of concentration is mol.dm -3, which is also written as M (stated as molar) A concentrated acid or base contains a large quantity of solute (acid or base) per volume of solution. A dilute acid or base contains a small quantity of solute (acid or base) per volume of solution. Calculations When a solution is diluted, the number of moles of the original solution stays the same even though the volume of the solution changes. The following equation is used. c 1 V 1 = c 2 V 2 c 1 = concentration of solution 1 (mol.dm -3 ) c 2 = concentration of solution 2 (mol.dm -3 ) V 1 = volume of solution 1 (dm 3 ) V 2 = volume of solution 2 (dm 3 ) For a titration reaction the following equation is used. n a n b = c av a c b V b n a = number of moles of acid (mol) n b = number of moles of base (mol) c a = concentration of acid (mol.dm -3 ) c b = concentration of base (mol.dm -3 V a = volume of acid (dm 3 ) V b = volume of base (dm 3 ) ph calculations The ph of a solution is an indication of the acidity or alkalinity of a solution. It is the negative logarithm of the hydronium ion concentration in a solution. ph = = log H + If the ph is known, [H + ] is found by: [H + ] = concentration of H + or H 3 O + ions H + = 10 ph Water undergoes auto-ionisation or auto-protolysis, in which a proton is transferred from one water molecule to another. This can be shown in the following equation: H 2 O l + H 2 O l H 3 O + aq + OH (aq) From this equation we get the equilibrium constant for this reaction, which is known as the dissociation constant for water: K w = [H 3 O + ][OH - ]

2 At 25 o C, K w = H 3 O + OH = , which means in a neutral solution that [H 3 O + ] = [OH - ] = 1 x 10-7 mol.dm -3 Therefore in aqueous solutions (at 25 o C): Hydrolysis [H 3 O + ][OH - ] = 1 x When an acid and a base react together one of the products formed is a salt. The salt will be neutral if a strong base and strong acid are reacted together. However, if the acid and base are not of comparable strengths then the salt formed will either be acidic or basic, depending on how it dissolves in water. Hydrolysis is the ability of the ions to react with the water molecules, thus altering the ph. Salt in water Salt of strong acid and strong base Salt of strong acid and weak base Salt of weak acid and strong base Example NaCl (HCl& NaOH) NH 4 Cl (HCl& NH 3 ) CH 3 COONa (CH 3 COOH & NaOH) ph in aqueous solution ph = 7 ph < 7 (acidic) ph > 7 (basic) Example: Consider the salt NH 4 Cl. This salt is made from the reaction between NH 3 (weak base) and HCl (strong acid). The Cl - ions will not react with the water molecules in the solution. But the NH 4 + ions will react with water according to the following equation: NH H 2 O NH 3 + H 3 O + Thus an excess of H 3 O + ions are created and the solution will be acidic. Multiple Choice Questions Question 1 Which ONE of the following is a CORRECT description for a 0,1 mol.dm -3 hydrochloric acid solution? A. Dilute strong acid B. Dilute weak acid C. Concentrated weak acid D. Concentrated strong acid Question 2 Which ONE of the following represents the products formed during the hydrolysis of ammonium chloride? A. NH 3 (aq) and H 3 O + (aq) B. NH + 4 (aq) and Cl - (aq) C. HCl(aq) and OH - (aq) D. Cl - (aq) and H 3 O + (aq)

3 Question 3 Consider the following reaction equilibrium: NH 3 g + H 2 O l NH 4 + aq + OH (aq) The two Bronsted-Lowry bases in the reaction equation are: A. NH 3 and H 2 O B. NH + 4 and OH - C. + H 2 O and NH 4 D. NH 3 and OH - Question 4 Water undergoes auto-ionisation. During this process A. a proton is transferred from one water molecule to another B. water molecules act as proton donors only C. water molecules act as proton acceptors only D. the ph of water will decrease Question 5 A small quantity of concentrated hydrochloric acid is gradually added to 1 dm 3 of distilled water at 25 o C. After testing the resultant solution, it is found that the value of K w, [H 3 O + ] and [OH - ] in mol.dm -3 are: A. K w = [H 3 O + ] < 10-7 [OH - ] > 10-7 B. K w < [H 3 O + ] < 10-7 [OH - ] < 10-7 C. K w = [H 3 O + ] > 10-7 [OH - ] < 10-7 D. K w = [H 3 O + ] = 10-7 [OH - ] = 10-7 Section B: Practice Questions Question 1 (Taken from Eastern Cape Paper 2 HG November2000) In an acid-base reaction, 500 cm 3 of a solution of sodium hydroxide is completely neutralised by 680 cm 3 of a solution of sulphuric acid. The equation for the reaction is H 2 SO 4 aq + 2NaOH aq Na 2 SO 4 (aq) + 2H 2 O(l) The ph of the base solution before any acid is added is 13,80 at 25 o C Show by calculation that the concentration of the NaOH solution will be 0,631 mol.dm -3. (4) 1.2. What is the difference between a strong base and a concentrated base? (2) 1.3. Calculate the mass of salt used to prepare the base solution. (4) 1.4. Calculate the number of moles of H 3 O + (aq) effectively used in the neutralisation process. (4) Question 2 (Taken from Northern Cape Paper 2 HG November 2000) When 500 cm 3 diluted hydrochloric acid of concentration 0,25 mol.dm -3 is added to 500 cm 3 of sodium hydroxide, the temperature of the solution rises and the ph changes to Classify this reaction as exothermic or endothermic. (1) 2.2. Calculate the final concentration of the H + ions (3) the initial concentration of the sodium hydroxide solution. (8) [14]

4 Question 3 (Taken from Northern Province Paper 2 HG November 2000) A 20 cm 3 solution of oxalic acid (COOH) 2 is titrated against a 0,25 mol.dm -3 solution of NaOH of which the volume was 28 cm 3. The unbalanced equation for the reaction is: COOH 2 aq + NaOH aq COONa 2 aq + H 2 O(l) 3.1. Balance the equation. (2) 3.2. Calculate the concentration of the oxalic acid. (4) 3.3. The salt, sodium oxalate (COONa) 2 reacts with water. What name is given to this reaction? (1) 3.4. Will the final solution for this reaction be acidic, basic or neutral? Give a reason for your answer. (4) Question 4 (Taken from KwaZulu Natal Paper 2 HG November 2000) A 0,5 dm 3 solution is made up by dissolving 4,0 g of sodium hydroxide in water Calculate the ph of this solution. (8) cm 3 of the above solution is neutralised by adding 40 cm 3 of dilute sulphuric acid solution. Calculate the concentration of the dilute sulphuric acid. (5) 4.3. The dilute sulphuric acid solution in 4.2 was prepared by adding 10 cm 3 of concentrated acid to 490 cm 3 of distilled water. Calculate the concentration of the concentrated sulphuric acid solution. (4) Section C: Solutions Multiple Choice Questions 1. A 2. A 3. D 4. A 5. C Practice Questions Question H + = 10 ph = 10 13,80 = 1, K w = H + OH = , OH = OH = 0,631 NaOH = OH = 0,631 mol. dm 3 (4) 1.2. A strong base is a base that has a high percentage ionisation in water A concentrated base contains a large number of moles of base per unit volume of solution. (2) 1.3. M(NaOH) = = 40 g.mol -1 c = m MV 0,631 = m 40 0,5 m = 0, ,5 = 12,62 g (4)

5 1.4. c NaOH = n V 0,631 = n 0,5 n = 0,631 0,5 = 0,316 mol 0,316 mol H 3 O + is needed to neutralise 0,316 mol NaOH (4) Question Exothermic (1) 2.2. H + = 10 ph = 10 2,3 = 5, mol. dm 3 (3) 2.3. Initial n(hcl) c = n V 0,25 = n 0,5 n = 0,25 0,5 = 0,125 mol Final n(hcl) c = n V 5, = n = 5, = 5, mol n (0,5 + 0,5) n HCl neutralised = 0,125 5, = 0,12 mol HCl + NaOH NaCl + H 2 O 0,12 mol NaOH reacts with 0,12 mol HCl Question 3 c NaOH = n V = 0,12 0,5 = 0,24 mol. dm 3 (8) 3.1. COOH 2 aq + 2NaOH aq COONa 2 aq + 2H 2 O l (2) 3.2. n a n b = c av a c b V b 1 2 = c a 0,02 0,25 0,028 c a = 0,25 0,028 (2)(0,02) = 0,175 mol. dm 3 (4) 3.3. Hydrolysis (1)

6 3.4. Basic Question 4 COONa 2 H 2 O (COO ) 2 + 2Na + (COO ) 2 + 2H 2 O COOH 2 + 2OH The OH - formed during hydrolysis causes the solution to become basic (4) 4.1. M(NaOH) = = 40 g.mol -1 c = m MV = ,5 = 0,2 mol. dm 3 [OH - ] = [NaOH] = 0,2 mol.dm -3 K w = [H + ][OH - ] ph = -log [H + ] = -log (5 x ) = [H + ](0,2) [H + ] = 5 x mol.dm -3 = 13,3 (8) 4.2. H 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O n a n b = c av a c b V b 1 2 = c a 0,04 0,2 0,02 c a = 0,2 0,02 (2)(0,04) = 0,05 mol. dm 3 (5) 4.3. c 1 V 1 = c 2 V 2 0,05 0,5 = c 2 0,01 c 2 = 2,5 mol. dm 3 (4)