Q. Antacid tablets should normally be chewed before they are swallowed. Why? A. Antacid tablets are normally chewed to provide a larger surface area for faster reaction with stomach acids. Q. A laboratory test to determine how much hydrochloric acid is neutralised by a brand of antacid does not give a complete picture of its effectiveness in the stomach. What other factors might be important? A. Other factors to consider when deciding an antacid s effectiveness include the neutralising action of the antacid over a prolonged period (30 minutes, for example), and whether or not the antacid upsets the acid balance in the stomach. The presence and nature of food in the stomach may also affect the neutralisation reaction. Furthermore, some brands claim to have a coating action on the stomach wall which might be unrelated to the neutralising action. Q3. Examine the range of antacids shown in Figure 4. on page 37 of the student book. If you were to choose one of these from all the others, what features, apart from its ability to neutralise acid, might influence your choice? A3. The consumer may have a preference for tablets, gels or solutions. He or she could also be influenced by price, attractiveness of packaging and whether or not the medication can be conveniently carried. Q4. Many antacids fizz when dissolved in a glass of water. One such brand lists among its ingredients, sodium hydrogen carbonate and citric acid. Write an ionic equation for the reaction between HCO 3 (aq) and H 3 O + (aq) responsible for the fizz. A4. HCO 3 (aq) + H 3 O + (aq) CO (g) + H O(l) Q5. In each of the following equations: i identify the acids and bases ii name the conjugate acid base pairs a NH 3 (aq) + H O(l) NH + 4 (aq) + OH (aq) b HSO 4 (aq) + H O(l) H 3 O + (aq) + SO 4 (aq) c NH + 4 (aq) + S (aq) NH 3 (aq) + HS (aq) d CH 3 COO (aq) + H 3 O + (aq) H O(l) + CH 3 COOH(aq) Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd)
A5. (Acid is listed first.) a i H O; NH 3 ii b i NH + 4, NH 3 and H O, OH - HSO 4 ; H O ii HSO 4, SO 4 and H 3 O +, H O c i NH + 4 ; S ii NH + 4, NH 3 and HS, S d i H 3 O + ; CH 3 COO ii CH 3 COOH, CH 3 COO and H 3 O +, H O Q6. The graphs in Figure 4.7 show the ph curves for titrations involving combinations of acids and bases of various strengths. You have a choice of phenolphthalein and methyl orange indicator. Phenolphthalein changes colour over a ph range 8. to 0.0. Methyl orange changes colour between ph 3. and 4.4. Decide which indicator(s) would be suitable to identify the equivalence point for each reaction. Provide reasons for your selections. a b c d Figure 4.7 Change in ph during a titrations of: a a strong acid with a strong base; b a strong acid with a weak base; c weak acid with a strong base; d weak acid with a weak base. Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd)
A6. a The equivalence point occurs in the range ph 3 to ph. Both indicators will change colour over this ph range. Both indicators will provide a sharp end point, i.e. they will change colour at the equivalence point with the addition a small volume, drop, of acid. b The equivalence point occurs in the ph range 3 to 7. Methyl orange provides a sharper end point over this ph range. c The equivalence point occurs in the ph range 7 to. Phenolphthalein provides the sharper end point. d Both indicators will provide a broad end point and neither would be suitable. Q7. The ethanoic acid content of white vinegar was determined by titrating a 0.00 ml aliquot of the vinegar with 0.995 M sodium hydroxide solution. The phenolphthalein indicator changed permanently from colourless to pink when 0.34 ml of sodium hydroxide solution was added from the burette. a Write an equation for the reaction. b Calculate the amount of sodium hydroxide, in mol, used in the titration. c Calculate the amount of ethanoic acid, in mol, used in the titration. d Calculate the concentration of ethanoic acid in the vinegar. A7. a Step Write a balanced equation. CH 3 COOH(aq) + NaOH(aq) CH 3 COONa(aq) + H O(l) b Step Calculate the amount of NaOH used in the titration, using n c V. n(naoh) 0.995 M 0.0034 L 0.004 mol c Step 3 Use the ratio of amounts of substances to calculate the amount of CH 3 COOH that was present in the 0.00 ml aliquot. From the equation in part a, mol CH 3 COOH reacts with mol NaOH. n(ch 3 COOH) n(naoh) n(ch 3 COOH) in 0.00 ml aliquot 0.004 mol d Step 4 Calculate the concentration of the CH 3 COOH in the 0.00 ml aliquot which is the same as the concentration in the vinegar, to the correct number of significant figures. c(ch 3 COOH) 0.004 mol 0.0000 L.0 M Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 3
Q8. The ammonia content of cloudy ammonia was determined first diluting a 5.0 ml sample to 50 ml in a volumetric flask. A 0.0mL aliquot of this solution was titrated with 0.0987 M hydrochloric acid. The volume of the acid used was.8 ml. The equation for the reaction is: NH 3 (aq) + HCl(aq) NH 4 Cl(aq) a Calculate the amount of hydrochloric acid, in mol, used in the titration. b Calculate the amount of ammonia, in mol, used in the titration. c Calculate the concentration of ammonia in the diluted solution used in the titration. d Calculate the concentration of ammonia in the original sample. A8. Remember to give the answer for each part with the correct number of significant figures, but keep all digits in your calculator for further calculations. a Step Calculate the amount of HCl used in the titration. n(hcl) 0.0987 M 0.08 L 0.008966 mol.9 0 3 mol b Step Write a balanced equation. NH 3 (aq) + HCl(aq) NH 4 Cl(aq) Step 3 Use the ratio of amounts of substances to calculate the amount of NH 3 that was present in the 0.00 ml aliquot. From the equation, mol NH 3 reacts with mol HCl. n(nh 3 ) n(hcl) n(nh 3 ) in 0.00 ml aliquot 0.008966 mol.9 0 3 mol c Step 4 Calculate the amount of NH 3 in the diluted solution. 0.008966 mol c(nh 3 ) in the diluted solution 0.000 L 0.094583 M 0.09 M d Step 5 Calculate the amount of NH 3 in 50 ml of the dilute solution. n c V n(nh 3 ) in the diluted solution 0.094583 M 0.50 L 0.073645 mol As this equals the amount of NH 3 in 5.0 ml of the original concentrated NH 3 solution, the concentration can be calculated. 0.073645 mol c(nh 3 ) in original solution 0.05 L.094583 M.09 M Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 4
Q9. A 0.4376 g aspirin tablet was heated gently with 50.00 ml of 0.96 M sodium hydroxide solution. The aspirin reacts according to the equation: C 6 H 4 (OCOCH 3 )COOH(aq) + NaOH(aq) C 6 H 4 (OH)COONa(aq) + CH 3 COONa(aq) + H O(l) After cooling, the resulting solution was titrated against 0.98 M hydrochloric acid to determine the amount of unreacted sodium hydroxide. A titre of 8.64 ml was obtained. Calculate: a the amount, in mol, of sodium hydroxide initially added to the aspirin b the amount of hydrochloric acid used c the amount of sodium hydroxide in excess after reaction with the aspirin d the amount of sodium hydroxide that reacted with the aspirin e the amount of aspirin in the tablet f the percentage, by mass, of aspirin in the tablet. A9. a n(naoh) initial 0.96 M 0.05000 L 0.00980 mol (three significant figures) b Step Write the balanced equation for the titration of HCl against the excess NaOH. HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) Step Calculate the amount of HCl needed to neutralise the excess NaOH. n(hcl) 0.98 M 0.0864 L 0.00555 mol (three significant figures) c From the equation, mol of NaOH reacted with mol of HCl. Calculate the amount of NaOH that reacted with the HCl, which was the amount of NaOH in excess after the aspirin reaction. n(naoh) excess n(hcl) n(naoh) excess 0.00555 mol d n(naoh) reacted n(naoh) initial n(naoh) excess (0.00980 0.00555) mol 0.0045 mol e From the equation for the reaction between aspirin and NaOH, mol of aspirin reacts with mol of NaOH. n(aspirin) n(naoh) n(aspirin) 0.0045 mol 0.003 mol f Step Calculate the mass of aspirin. m(aspirin) 0.003 mol 80 g mol 0.3834 g Step Find the percentage of aspirin in the tablet. % aspirin 0.3834 00 0.4376 87.6% (three significant figures) Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 5
Chapter review Q0. Write full equations for these acid base reactions: a nitric acid is added to sodium hydroxide solution b sulfuric acid is added to potassium hydroxide solution c hydrochloric acid is added to ammonia solution d ethanoic acid solution is added to potassium hydroxide solution. A0. a HNO 3 (aq) + NaOH(aq) NaNO 3 (aq) + H O(l) b H SO 4 (aq) + KOH(aq) K SO 4 (aq) + H O(l) c HCl(aq) + NH 3 (aq) NH 4 Cl(aq) d CH 3 COOH(aq) + KOH(aq) CH 3 COOK(aq) + H O(l) Q. Write ionic equations for the reactions in Question 0. A. a H + (aq) + OH (aq) H O(l) b H + (aq) + OH (aq) H O(l) c H + (aq) + NH 3 (aq) NH + 4 (aq) d CH 3 COOH(aq) + OH (aq) CH 3 COO (aq) + H O(l) Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 6
Q. What mass of sodium sulfate is produced when 5.0 ml of 0.00 M sulfuric acid is added to 0.0 ml of 0.5 M sodium hydroxide solution? A. Step Step Step 3 Step 4 Write a balanced equation. H SO 4 (aq) + NaOH(aq) Na SO 4 (aq) + H O(l) To determine which reactant is in excess, calculate amount of each reactant divided by their respective coefficient. The smallest amount is the limiting reactant and the one from which to calculate amount of product formed. The other is the excess reactant. Note: These calculations can only be used to determine the excess reactant. Continue the calculations, using original data. n(h SO 4 ) 0.05 L 0.00 M coefficient(h SO 4 ) 0.005 mol n(naoh) 0.000 L 0.5 M coefficient(naoh) 0.005 mol Hence NaOH is the limiting reactant. From the equation, mol of Na SO 4 is produced by mol of NaOH. n(na SO 4 ) n(naoh) n(naoh) n(na SO 4 ) 0.000 L 0.5 M 0.005 mol Calculate the mass of Na SO 4 to the correct number of significant figures. m(na SO 4 ) 0.005 mol 4.04 g mol 0.306 g 0. g Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 7
Q3. What volume of 0.00 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide? A3. The NaOH and KOH react independently with the H SO 4. Treat them as separate reactions and write two balanced equations. Add the volumes of H SO 4 from each reaction to find the total volume needed to neutralise the solution. Step For the NaOH solution, write a balanced equation. NaOH(aq) + H SO 4 (aq) Na SO 4 (aq) + H O(l) Step Calculate amount of NaOH. n(naoh) 0.500 g 39.98 g mol 0.0506 mol Step 3 From the equation, mol of HSO4 is produced from mol of NaOH. n(h SO 4 ) n(naoh) n(h SO 4 ) 0. 0506 mol 0.00653 mol Step 4 Calculate the volume of H SO 4 needed to neutralise the NaOH. V(H SO 4 ) 0.006533 mol 0.00 M 0.0653 L Step 5 For the KOH solution, write a balanced equation. KOH(aq) + H SO 4 (aq) K SO 4 (aq) + H O(l) Step 6 Calculate amount of KOH. n(koh) 0.800 g 56.08 g mol 0.046 mol Step 7 From the equation, mol of H SO 4 is produced from mol of KOH. n(h SO 4 ) n(koh) n(h SO 4 ) 0.046 mol 0.0079 mol Step 8 Calculate the volume of H SO 4 needed to neutralise the KOH. V(H SO 4 ) 0.0079 mol 0.00 M 0.079 L Step 9 Calculate the total volume of H SO 4 needed to neutralise the KOH and the NaOH. V(H SO 4 ) (0.079 + 0.0653) L 0.338 L 0.34 L Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 8
Q4. The cleaning agent in a household window cleaner is an aqueous solution of ammonia. Draw a flow chart to show how you would perform a titration to find the concentration of ammonia in the window cleaner. A4. Place a standard solution of hydrochloric acid in a burette. Record the initial volume. Place an aliquot of the window cleaner in a conical flask. 3 Add two or three drops of indicator to the cleaner. (Methyl orange is a suitable indicator.) 4 Titrate the window cleaner with the hydrochloric acid. Record the volume of the acid used to reach the end point. 5 Repeat steps 4 to obtain three concordant titres (titres within 0. ml). In practice, it may be necessary to dilute the window cleaner, using a pipette and volumetric flask, in order to obtain titres that are not excessively large. Q5. Read the description of the analysis of concrete cleaner on page 4 again. What safety instructions would you give to a student intending to perform an analysis such as this? A5. Hydrochloric acid is very corrosive and its vapour irritates the skin, eyes and respiratory system. You should instruct the student to wear safety glasses and avoid skin contact when dealing with this substance. A pipette filler should always be used when dispensing aliquots of this liquid. Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 9
Q6. The concentration of sodium hydroxide in waste water from an alumina refinery was found by titrating 0.00 ml aliquots of the water against 0.50 M hydrochloric acid, using phenolphthalein as indicator. The average titre of several titrations was.40 ml. a Why is an indicator used? b Write an equation for the reaction that occurred. c What was the molarity of NaOH in the waste water? d What mass of NaOH would be present in 00 L of the waste water? A6. a An indicator is necessary to detect the equivalence point as both the reactants and products in the reaction are colourless. b NaOH(aq) + HCl(aq) NaCl(aq) + H O(l) c Step n(hcl) 0.50 M 0.04 L 0.007 mol Step From the equation, mol NaOH reacts with mol HCl. n(naoh) n(hcl) n(naoh) 0.007 mol Step 3 Calculate the concentration of NaOH. c(naoh) 0.007 mol/0.000 L 0.0855 M d Step Calculate the mass of NaOH present in L to the correct number of significant figures. c(naoh) 0.0855 mol L m(naoh) (0.0855 mol 39.98 g mol ) g L 3.483 g L Step Convert this to the mass of NaOH present in 00 L to the correct number of significant figures. m(naoh) 3.483 g L (3.483 00) g in 00 L of water 34.83 g 34 g Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 0
Q7. A 4.7 ml volume of a hydrochloric acid solution is required to react completely with 0.0 ml of 0.6 M sodium carbonate solution. a Write an equation for the reaction. b Calculate the concentration of the HCl, in mol L. A7. a HCl(aq) + Na CO 3 (aq) NaCl(aq) + H O(l) + CO (g) b Step Calculate the amount of Na CO 3. n(na CO 3 ) 0.6 M 0.000 L 0.04 mol Step From the equation, mol HCl reacts with mol of Na CO 3. n(hcl) n(na CO3 ) n(hcl) 0.04 mol 0.0448 mol Step 3 Calculate the concentration of HCl to the correct number of significant figures. c(hcl) 0.0448 mol/0.047 L 0.5733 M 0.573 M Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd)
Q8. A.0 g antacid tablet contains 80.0% by mass of Mg(OH) as the active ingredient. What volume of 0.500 M HCl could the antacid tablet neutralise? A8. Step Write a balanced equation. Mg(OH) (s) + HCl(aq) MgCl (aq) + H O(l) Step Calculate the mass of Mg(OH) in tablet. 80 m(mg(oh) ).0 g 00 0.960 g Step 3 Calculate the amount of Mg(OH). 0.960 g n(mg(oh) ) 58.36 g mol 0.0646 mol Step 4 From the equation, mol HCl reacts with mol Mg(OH). n(hcl) n(mg(oh) ) n(hcl) 0.0646 mol 0.039 mol Step 5 Calculate the volume of HCl required for neutralisation, to the correct number of significant figures. 0.039 mol V(HCl) 0.500 M 0.96 L 0 ml Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd)
Q9. Washing soda is added to hard water to allow soap to lather. A brand of washing soda contains partially hydrated sodium carbonate solid. A 0.300 g sample completely reacts with 0.0 ml of 0.50 M hydrochloric acid. a What mass of sodium carbonate was present? b Calculate the percentage by mass of sodium carbonate in the washing soda. A9. a HCl(aq) + Na CO 3 (aq) NaCl(aq) + H O(l) + CO (g) Step Calculate the amount of HCl. n(hcl) 0.50 M 0.000 L 0.00500 mol Step From the equation, mol Na CO 3 reacts with mol HCl. n(na CO3 ) n(hcl) 0.00500 mol n(na CO 3 ) 0.0050 mol Step 3 Calculate the mass of Na CO 3, to the correct number of significant figures. m(na CO 3 ) 0.0050 mol 05.99 g mol 0.64975 g 0.65 g b Express as a percentage to the correct number of significant figures. 0.650 g % Na CO 3 00% 0.300 g 88.33% 88.3% Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 3
Q0. A 50.0 ml sample of vinegar was diluted to 50.0 ml in a volumetric flask. A 0.00 ml aliquot of this solution required the addition of 7.98 ml of 0.34 M sodium hydroxide solution in order to be neutralised. a Write an equation for the neutralisation reaction. b What is the molarity of ethanoic acid in the original vinegar? c Express your answer to part b in g L. A0. a CH 3 COOH(aq) + NaOH(aq) CH 3 COONa(aq) + H O(l) b Step Calculate the amount of NaOH used in neutralisation. n(naoh) 0.34 M 0.0798 L 0.003749 mol Step From the equation, mol CH 3 COOH reacts with mol NaOH in the 0.00 ml aliquot. n(ch 3 COOH) n(naoh) n(ch 3 COOH) in 0.00 ml 0.003749 mol Step 3 Calculate the amount of CH 3 COOH in the 50.0 ml flask, remembering that only 0.00 ml was removed from the flask. 50.0 n(ch 3 COOH) in 50.0 ml 0.003749 mol 0.00 0.04687 mol Step 4 The amount of CH 3 COOH in the original 50.0 ml of concentrated vinegar is the same amount as in the 50.0 ml flask. Calculate the concentration of original vinegar solution. c(ch 3 COOH) in 50.00 ml sample 0.04687 mol/0.0500 L 0.9373 M 0.937 M c Convert mol L to concentration in g L, using m n M. c(ch 3 COOH) (0.9373 mol 60.05 g mol ) g L 56.69 g L 56.3 g L Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 4
Q. A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus: a the pipette used to deliver the aliquot of sodium carbonate solution b the flask containing the aliquot c the burette. A. a higher b unchanged c lower Q. In order to standardise a solution of hydrochloric acid, a student titrated the solution against 0.00 ml aliquots of a standard solution of sodium carbonate. Methyl orange indicator was used to identify the end point of the reaction: HCl(aq) + Na CO 3 (aq) NaCl(aq) + H O(l) + CO (g) The sodium carbonate solution had been prepared by dissolving.36 g of anhydrous Na CO 3 in water and making the solution up to 50.0 ml in a volumetric flask. The titres recorded were.56 ml, 0.98 ml, 0.96 ml and.03 ml. a What value for the titre of hydrochloric acid solution should the student use in the calculation of the acid concentration? Explain your answer. b What is the molarity of the hydrochloric acid solution? c Calculate the concentration of the HCl, in mol L. A. a 0.99 ml. Use the average of the last three titres. The student appears to have passed the end point with the first titre. b Step Write a balanced equation. HCl(aq) + Na CO 3 (aq) NaCl(aq) + H O(l) + CO (g) Step Calculate the concentration of the Na CO 3 standard solution. c(na CO 3 ) m MV.36 g 05.96 g mol 0.500 L 0.046 659 M 0.046 66 M (four significant figures) Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 5
c Step From the equation, mol of HCl reacts with mol of Na CO 3 in the titration. n(hcl) n(na CO3 ) n(hcl) 0.04666 M 0.0000 L 0.0086636 mol Step Calculate the concentration of HCl. c(hcl) 0.00 6636 mol/0.00 99 L 0.088 98 M 0.088 9 M Q3. A solution of a metal carbonate with the formula M CO 3 was prepared by dissolving.80 g of the solid in 50 ml of water. 0.00 ml aliquots of this solution were then titrated with 0.50 M H SO 4, using methyl orange indicator. The average titre was 0.8 ml. a Write an equation for the reaction between the metal carbonate and sulfuric acid. b What amount of sulfuric acid, in mol, was needed to reach the end point of the titration? c What amount of the metal carbonate was present in each 0.00 ml aliquot? d What amount of metal carbonate was present in the original sample? e Determine the molar mass of the metal carbonate. f What is the identity of the metal M? A3. a H SO 4 (aq) + M CO 3 (aq) M SO 4 (aq) + H O(l) + CO (g) b n(h SO 4 ) 0.50 M 0.008 L 0.00 6 mol (three significant figures) c From the equation, mol M CO 3 reacts with mol H SO 4 in the titration. n(m CO 3) n(h SO 4 ) n(m CO 3 ) in 0.00 ml 0.00 6 mol d Calculate the amount of M CO 3 in the 50.0 ml flask, remembering that only 0.00 ml was removed from the 50.0 ml flask for the titration. 50.0 n(m CO 3 ) in 50.0 ml 0.00 6 mol 0.00 e 0.00 5 mol 0.003 mol (three significant figures) The amount of M CO 3 in the.80 g is the same amount as is in the 50.0 ml flask. Calculate the molar mass of M CO 3, using M n m..80 g M(M CO 3 ) 0.005 mol 38.7 g mol 38 g mol (three significant figures) Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 6
f By calculating molar mass of CO 3, determine which element is M. M(CO 3 ) 60 g mol Q4. M(M) M (M CO ) M (CO 3 3 38 60 39 g mol This is the molar mass of potassium. ) Sketch the ph curve that would be obtained when a strong base is titrated with a weak acid. A4. Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 7
Q5. A 50 ml volume of 0.0 M nitric acid is mixed with 60 ml of 0.0 M calcium hydroxide solution. What volume of 0.050 M sulfuric acid is required to neutralise the mixture? A5. Step Write a balanced equation for the first reaction occurring. HNO 3 (aq) + Ca(OH) (aq) Ca(NO 3 ) (aq) + H O(l) Step There is no need to determine which reactant is in excess. You are told that H SO 4 is used to neutralise the solution, which clearly indicates that the base, Ca(OH), is the excess reactant. From the equation, mol of Ca(OH) reacts with mol of HNO 3. n(ca(oh) ) n(hno3) n(hno n(ca(oh) ) reacted 3 ) 0.050 L 0.00 M 0.0050 mol Step 3 Calculate the amount of excess Ca(OH). n(ca(oh) ) excess n(ca(oh) ) initial n(ca(oh) ) reacted 0.0060 mol 0.0050 mol 0.0035 mol Step 4 Write a balanced equation for the second (neutralisation) reaction. Ca(OH) (aq) + H SO 4 (aq) CaSO 4 (aq) + H O(l) Step 5 Knowing that 0.0035 mol of Ca(OH) needs to be neutralised, calculate the amount of H SO 4 needed for the neutralisation. From the equation, mol of H SO 4 reacts with mol of Ca(OH). n(h SO 4 ) n(ca(oh) ) n(h SO 4 ) n(ca(oh) ) 0.0035 mol Step 6 Calculate the volume of H SO 4. 0.0035 mol V(H SO 4 ) 0.050 M 0.070 L 70 ml (two significant figures) Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 8
Q6. Lawn fertiliser contains ammonium ions (NH + 4 ). A.34 g sample of lawn fertiliser was dissolved in water to make a 50.0 ml solution. A 0.00 ml aliquot of this solution was added to a flask containing 0.00 ml of 0.0 M sodium hydroxide solution. The flask was heated until the reaction: NH + 4 (aq) + OH (aq) NH 3 (aq) + H O(l) was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.3 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 ml had been added. Calculate: a the amount, in mol, of HCl used in the titration b the amount of NaOH in excess after reaction with the fertiliser c the amount of NaOH that reacted with the NH + 4 ions d e the amount of NH + 4 ions in the.34 g fertiliser sample the percentage by mass of nitrogen in the fertiliser, assuming nitrogen is only present as ammonium ions. A6. a Step Write a balanced equation for the titration. HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) Step Calculate the amount of HCl used in the titration. n(hcl) 0.3 M 0.009 97 L 0.00 9 mol 0.00 3 mol (three significant figures) b From the equation, mol of NaOH reacts with mol of HCl. This is the amount of NaOH in excess after reaction with the fertiliser. n(naoh) excess n(hcl) n(naoh) excess n(hcl) 0.00 3 mol c Step Write the balanced equation for the reaction of 0.00 ml of fertiliser and NaOH. NH + 4 (aq) + OH (aq) NH 3 (aq) + H O(l) Step Calculate the amount of NaOH added initially. n(naoh) initial 0.0 M 0.000 L 0.00 044 mol Step 3 Calculate the amount of NaOH that reacted with the fertiliser in the 0.00 ml aliquot. n(naoh) reacted n(naoh) initial n(naoh) excess (0.00 044 0.00 9) mol 0.000 95 mol Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 9
d Step From the equation, mol of fertiliser reacted with mol of NaOH. + n(nh 4 ) n(oh ) n(nh + 4 ) in 0.00 ml 0.00095 mol Step Calculate the amount of fertiliser, NH + 4, in the 50.0 ml flask, remembering that only 0.00 ml was removed from the 50.0 ml flask for the titration. This is the same amount of NH + 4 as is present in.34 g of fertiliser. n(nh + 50.0 4 ) in 50.0 ml 0.00095 0.00 mol 0.04 mol (three significant figures) e Step Calculate the mass of N in.34 g of fertiliser. n(n) n(nh + 4 ) m(n) 0.044 mol 4.0 g mol 0.603 g Step Convert to a percentage by mass of N in the fertiliser. % N 0.603 00%.34 3.0% (three significant figures) Q7. The drain cleaner Drainol contains 3% m/v NaOH. The original solution is diluted, by taking 0 ml and making it up to 500 ml with distilled water. a What is the molarity of the original solution? b Concentrated solutions of NaOH are not used in a titration. Give two reasons why. c The concentration of the diluted NaOH is determined by titration. Give a suitable reactant and indicator to use in the titration with NaOH. d Standard NaOH solutions cannot be prepared directly from the solid. Explain why. A7. a Determine the moles of NaOH in L. 3% m/v NaOH 3 g NaOH in 00 ml solution 30 g NaOH L b c d n(naoh) 30 g 3.5 M 40.0 g mol Any two of the following (reduce errors, safety, cost, solubilising silicate). - Measurement errors are higher if solutions are concentrated. - Concentrated sodium hydroxide can cause severe burns. - Concentrated solutions are expensive as more reagent is used. - Concentrated sodium hydroxide will solubilise silicate (glass). This will reduce the accuracy of the glassware. Any strong acid and acid base indicator, for example hydrochloric acid and methyl orange. If a weak acid is chosen such as acetic acid, phenolphthalein should be selected as the indicator. Sodium hydroxide pellets readily absorb water and carbon dioxide from the atmosphere so cannot be readily obtained as a pure solid. Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd) 0
Q8. An impure sample of limestone, mainly calcium carbonate, was analysed by using a back titration. Approximately g of the finely powdered limestone was weighed accurately into a conical flask. An excess of HCl, exactly 50.00 ml, was added to the limestone. The mixture was stirred for 5 min with a magnetic stirrer to allow the reaction to be completed. The mixture was then titrated with a standard solution of NaOH and the following results were obtained: - Mass of watch glass 8.7954 g - Mass of limestone and watch glass 9.8460 g - Concentration of standard NaOH solution 0.0489 M - Titration value of NaOH obtained.3 ml - Concentration of HCl 0.395 M a Write balance equations for the two reactions that occur. b Determine the moles of HCl in excess after the reaction with the limestone. c Calculate the total moles of HCl added to the limestone. d How many moles of HCl reacted with the limestone? e Calculate the number of moles of calcium carbonate in the limestone. f g What is the percentage of calcium carbonate in the limestone. In this experiment the whole sample of limestone was used in one titration. How could the precision of the titration have been improved? A8. a HCl(aq) + CaCO 3 (s) CaCl (aq) + H O(l) + CO (g) HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) b Calculate the number of moles of sodium hydroxide used..3 n(naoh) 0.0489 M 0.00 09 mol 000 L Use mole ratios in the balanced equation to determine number of moles of HCl(aq). n(hcl(aq)) n(naoh) 0.00 09 mol 50.00 ml c Total n(hcl) 0.395 M 0.09 75 mol 0.098 mol 000 ml d n(hcl) reacted with limestone 0.09 75 0.00 09 0.08 66 mol 0.087 mol 0.0866 mol e n(caco 3 ) 0.009 330 mol m(caco 3 ) 0.009 330 mol 00. g mol 0.9339 g f m(limestone ) 9.8460 8.7954.0506 g 0.9339 % calcium carbonate 00% 88.9%.0506 g Dissolve an accurately known mass of limestone, for example 0 g, in exactly 500 ml of hydrochloric acid. Take aliquots of the solution and titrate until concordant results are obtained. Copyright Pearson Australia 00 (a division of Pearson Australia Group Pty Ltd)