JK-TK/Newdata huggingface dataset:
(i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3.
The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3. The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution. Examples 2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. 0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. “2M” is more concentrated than“0.02M”. Preparation of molar solution
26 26 Procedure Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask. Using a wash bottle add about 200cm3 of distilled water. Stopper the flask. Shake vigorously for three minutes. Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved. Add more water slowly upto exactly the 250 cm3 mark. Sample questions 1.Calculate the number of moles of sodium hydroxide pellets present in: (i) 4.0 g. Molar mass of NaOH = (23 + 16 + 1) = 40g Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles Molar mass 40 (ii) 250 cm3 solution in the volumetric flask.
x 10-3moles 1000 1000 Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g 3. Calculate the volume required to dissolve : (a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3 Molarity 0.8 (ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.8M x 312.5cm3 = C2 x (312.5 + 100) C2 = 0.8M x 312.5cm3 = 0.6061M
31 31 412.5 (b)(ii) 0.01M solution containing 0.01moles of sodium hydroxide solution . Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products.
Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products. Knowing the concentration/ volume of one reactant, the other can be determined from the relationship: M1V1 = M2V2 where: n1 n2 M1 = Molarity of 1st reactant M2 = Molarity of 2nd reactant V1 = Volume of 1st reactant V1 = Volume of 2nd reactant n1 = number of moles of 1st reactant from stoichiometric equation n2 = number of moles of 2nd reactant from stoichiometric equation Examples 1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0) Stoichiometric equation: MCO3(s) + 2HCl(aq) -> MCl2(aq) + CO2(g) + H2O(l) Method 1
32 32 M1V1 = M2V2 -> M1 x 5.0cm3 = 0.5M x 25.0cm3 n1 n2 1 2 => M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3 5.0 x 2 Method 2 Moles of HCl used = molarity x volume 1000 => 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles 1000 Mole ratio MCO3 : HCl = 1:2 Moles MCO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles 2 Molarity MCO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000 Volume 5 = 1.25M / 1.25 moledm-3 2.
0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B Moles of NaOH used = molarity x volume 1000 => 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles 1000 Mole ratio B : NaOH = 1:1
34 34 Moles B = 0.0044175 /4.4175 x 10-3moles Molar mass B = mass => 0.388 moles 0.0044175 /4.4175 x 10-3moles = 87.8324 gmole-1 X-COOH = 87.8324 where X is an alkyl group X =87.8324- 42 = 42.8324=43 By elimination: CH3 = 15 CH3CH2 = 29 CH3CH2 CH2 = 43 Molecula formula : CH3CH2 CH2COOH Molecule name : Butan-1-oic acid Molecular structure H H H O H C C C C O H H H H H 5. 10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer.
10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0) Equation for neutralization NaOH(aq) + HCl(aq) -> NaOH(aq) + H2O(l) Mole ratio NaOH(aq):HCl(aq)= 1:1 Moles of HCl = Molarity x volume => 0.5 x 85 = 0.0425 moles 1000 1000 Excess moles of NaOH(aq)= 0.0425 moles Equation for reaction with ammonium salt 2NaOH(aq) + (NH4) 2SO4(aq) -> Na 2SO4(aq) + 2NH3 (g)+ 2H2O(l) Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1 Total moles of NaOH = Molarity x volume => 0.8 x 250 = 0.2 moles 1000 1000 Moles of NaOH that reacted with(NH4) 2SO4 = 0.2 - 0.0425 = 0.1575moles Moles (NH4) 2SO4 = ½ x 0.1575moles = 0. 07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction.
07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point. Set up of titration apparatus The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings. To reduce errors, titration process should be repeated at least once more. The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.
The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided. Calculate the average volume of solution used 24.0 + 24.0 + 24.0 = 24.0 cm3 3 As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point. As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value . The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates. Bonus mark is awarded for averaged reading within 0.1 school value as Final answer. Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the: (i)relationship among the mole, molar mass, mole ratios, concentration, molarity. (ii) mathematical application of 1st principles. Very useful information which candidates forget appears usually in the beginning of the question paper as: “You are provided with…” All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point. Candidates are expected to use a non programmable scientific calculator. 37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask.
37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1 Sample results Titration table 1 Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution B used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution B used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre.
How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 = 0.1M Volume 20
38 38 (b)Sample Titration Practice 2 (Redox Titration) You are provided with: Acidified Potassium manganate(VII) solution A 0.1M of an iron (II)salt solution B 8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O solid C You are required to (i)standardize acidified potassium manganate(VII) (ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O. Procedure 1 Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears. Record your results to complete table 1. Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution.
Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution A were present in the average volume. Assume one mole of B react with five moles of B Mole ratio A : B = 1:5 => Moles of A = Moles of B = 2.5 x 10-3 moles = 5.0 x 10 -4 moles 5 5
39 39 (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 Volume 20 = 0.025 M /moles per litre /moles l-1 Procedure 2 Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1.
Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present inin the average titre. Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O.
Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O. 40 40 (N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0) Molar mass = mass perlitre = 8.5 = 8500 g Moles per litre 1.0 x 10 -3 moles NH4)2SO4FeSO4.xH2O = 8500 284 + 18x =8500 8500 - 284 = 8216 = 18x = 454.4444 18 18 x = 454 (whole number) (c)Sample Titration Practice 3 (Back titration) You are provided with: (i)an impure calcium carbonate labeled M (ii)Hydrochloric acid labeled solution N (iii)solution L containing 20g per litre sodium hydroxide. You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator.
You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M
41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette.
Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M
41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.
Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?