CHEMSHEETS.co.uk 14-Jul-12 Chemsheets A2 009 1
www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 0091
BRONSTED-LOWRY ACIDS & BASES Bronsted-Lowry acid proton donor (H proton) Bronsted-Lowry base proton acceptor (H proton)Bronsted-Lowry acid-base reaction reaction involving the transfer of a proton He.g.HClH donates H acidNaOH - NaCl H2Oe.g.HNO3 OH accepts H basedonates H acidNH3 NH4 NO3- NH3 accepts H baseTASK 1 – Bronsted-Lowry acids & basesIdentify the Bronsted-Lowry acid and base in each of the following reactions.acid- i)H2O NH3 OHii)H2O HCl H3Oiii)KOH HCOOH iv)CH3COOH HCl CH3COOH2 Clv)NH3 HCl vi)HCO3 OHvii)HCO3 H NH4 - ClHCOOK H2O --- -NH4Cl baseCO32- H2OCO2 H2Oviii) H2SO4 HNO3 www.CHEMSHEETS.co.ukHSO4- H2NO3 14-Jul-12Chemsheets A2 0092
pH OF STRONG ACIDS Monoprotic acid acid that releases one H ion per moleculee.g. HCl (hydrochloric acid), HNO3 (nitric acid), CH3COOH (ethanoic acid) Diprotic acid acid that releases two H ions per moleculee.g. H2SO4 (sulfuric acid), H2C2O4 (ethanedioic acid)Moles of H Moles of acidMoles of H Moles of acid3 moles of HNO30.1 moles of H2SO42 moles of HCl0.2 moles of HCl4 moles of H2SO40.08 moles of HNO30.3 moles of HNO30.08 moles of H2SO40.3 moles of H2SO40.35 moles of HClpH – log [H ]Definition of pHUseful rearrangement[H ] 10-pHALWAYS give pH to 2 DECIMAL PLACES[H ]0.00100pH2.5 x 10-41.502.753.304.5 x 10-1213.701.85-0.70Calculating the pH of a strong acid-3[H ] 0.500pH -log 0.500pH 0.30pH of 0.200 mol dm H2SO4?-3[H ] 2 x 0.200 0.400 (diprotic acid!)pH -log 0.400pH 0.40[HCl] with pH 1.70?[H ] 10 0.0200-3[HCl] 0.0200 mol dm[H2SO4] with pH 1.30?[H ] 10 0.0501-3[H2SO4] 0.0501 / 2 0.251 mol dmpH of 0.500 mol dm HNO3? www.CHEMSHEETS.co.uk -1.70 -1.3014-Jul-12Chemsheets A2 0093
Dilution of a strong acid33-3Calculate the pH of the solution formed when 100 cm of water is added to 50 cm of 0.100 mol dm HNO3. [H ] in original HNO3 solution 0.100 [H ] in diluted solution 0.100 x 50 0.0333150pH -log 0.0333 1.483-33Calculate the pH of the solution formed when 250 cm of 0.300 mol dm H2SO4 is made up to 1000 cm solution with water. [H ] in original H2SO4 solution 2 x 0.300 0.600 [H ] in diluted solution 0.600 x 250 0.1501000pH -log 0.150 0.82TASK 2 – pH of strong acids1)Calculate the pH of the following solutions.-3a) 0.2 mol dm HCl-3b) 0.05 mol dm HNO3-3c) 0.04 mol dm H2SO4-3d) 2.00 mol dm HNO32)Calculate the concentration of the following acids.a) HCl with pH 3.55b) H2SO4 with pH 1.70c) HNO3 with pH 1.30d) H2SO4 with pH -0.503)Calculate the pH of the solutions formed in the following way.33-33-3a) addition of 250 cm of water to 50 cm of 0.200 mol dm HNO33b) addition of 25 cm of water to 100 cm of 0.100 mol dm H2SO43-33c) adding water to 100 cm of 2.00 mol dm H2SO4 to make 500 cm of solution3-33d) adding water to 25 cm of 1.50 mol dm HCl to make 250 cm of solution4)Calculate the pH of the following solutions.-3a) 10 g dm HCl-3b) 20 g dm H2SO4-3c) 50 g dm HNO3-3d) 100 g dm H2SO4 www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 0094
THE IONIC PRODUCT OF WATER, KW The ionic product of water (Kw) -H2O ¾ H OH H endothermic-Kc [H ] [OH ][H2O] - Kc [H2O] [H ] [OH ] -As [H2O] is very much greater than [H ] and [OH ], then [H2O] is effectively a constant number Kc [H2O] a constant KwKw [H ] [OH-]The effect of temperature on the pH of water and the neutrality of waterAs the temperature increases, the equilibrium moves right to lower the temperature - [H ] and [OH ] increase Kw increases and pH increases - -However, the water is still neutral as [H ] [OH ] (and the definition of neutral is [H ] [OH ])Calculating the pH of water -In pure water, [H ] [OH ] 2 Kw [H ] [H ] KwCalculate the pH of water at 40 C when Kw 2.09 x 10-142-6mol dm . 2Kw [H ] -14 [H ] Kw (2.09 x 10 ) 1.45 x 10-7-7 pH -log (1.45 x 10 ) 6.84Calculate the pH of water at 100 C when Kw 51.3 x 10-142-6mol dm . . . . . . . . . www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 0095
THE pH OF STRONG BASES-e.g.NaOH, KOH, NH3-e.g.Ba(OH)2, Ca(OH)2Monobasic base base that releases one OH ionDibasic base base that releases two OH ionsCalculating the pH of a strong base-3pH of 0.200 mol dm NaOH?-[OH ] 0.200 -14[H ] Kw 10 5 x 10[OH ] 0.200 -14-14pH -log[H ] -log(5 x 10 ) 13.30-3pH of 0.0500 mol dm Ba(OH)2?-[OH ] 2 x 0.0500 0.100 -14[H ] Kw 10 1 x 10[OH ] 0.100 -13-13pH -log[H ] -log(1 x 10 ) 13.00 [KOH] with pH 12.70?[H ] 10-pH 10-12.70- 2.00 x 10-13-14[OH ] Kw 10 0.05 -13[H ] 2.00 x 10[KOH] 0.05 mol dm [Ba(OH)2] with pH 13.30?[H ] 10-pH 10-13.30--3 5.01 x 10-14-14[OH ] Kw 10 0.200 -14[H ] 5.01 x 10[Ba(OH)2] 0.100 mol dm-3Dilution of a strong base33-3Calculate the pH of the solution formed when 50 cm of water is added to 100 cm of 0.200 mol dm NaOH.-[OH ] in original NaOH solution 0.200-[OH ] in diluted solution 0.200 x 50 0.0667150 -14[H ] Kw 10 1.50 x 10[OH ] 0.0667-13-13pH -log (1.50 x 10 ) 12.823-33Calculate the pH of the solution formed when 50 cm of 0.250 mol dm KOH is made up to 250 cm solution with water.-[OH ] in original KOH solution 0.250-[OH ] in diluted solution 0.250 x 50 0.0500250 -14[H ] Kw 10 2.00 x 10[OH ] 0.0500-13-13pH -log (2.00 x 10 ) 12.70 www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 0096
TASK 3 – pH of strong bases1)Calculate the pH of the following solutions.-3a) 0.15 mol dm KOH-3b) 0.05 mol dm NaOH-3c) 0.20 mol dm Ba(OH)22)Calculate the concentration of the following acids.a) NaOH with pH 14.30b) Ba(OH)2 with pH 12.50c) KOH with pH 13.703)Calculate the pH of the solutions formed in the following way.33-33-3a) addition of 100 cm of water to 25 cm of 0.100 mol dm NaOH3b) addition of 25 cm of water to 100 cm of 0.100 mol dm Ba(OH)23-33c) adding water to 100 cm of 1.00 mol dm KOH to make 1 dm of solution4)Calculate the pH of the following solutions.-3a) 20 g dm NaOH-3b) 100 g dm KOH-3c) 1 g dm Sr(OH)2Reaction between a strong acid and a strong base1)2)3)4)5) Calculate moles HCalculate moles OH Calculate moles XS H or OH Calculate XS [H ] or XS [OH ]Calculate pH3-33-3Calculate the pH of the solution formed when 50 cm of 0.100 mol dm H2SO4 is added to 25 cm of 0.150 mol dm NaOH. mol H 2 x-mol OH 2550/1000 x 0.100 0.0100/1000 x 0.150 0.00375 XS mol H 0.0100 – 0.00375 0.00625 XS [H ] 0.00625 0.083375/1000 pH -log (0.0833) 1.08 www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 0097
3-33-3Calculate the pH of the solution formed when 25 cm of 0.250 mol dm H2SO4 is added to 100 cm of 0.200 mol dm NaOH. mol H 2 x-25100mol OH /1000 x 0.250 0.0125/1000 x 0.200 0.0200- XS mol OH 0.0200 – 0.0125 0.0075 XS [OH ] 0.0075 0.0600125/1000 XS [H ] Kw 10[OH ] 0.0600- -14-13 1.67 x 10-13pH -log (1.67 x 10 ) 12.78TASK 4 – strong acid strong base3-33-33-33-33-31)Calculate the pH of the solution formed when 20 cm of 0.100 mol dm HNO3 is added to 30 cm of 0.050 mol dmKOH.2)Calculate the pH of the solution formed when 25 cm of 0.150 mol dm H2SO4 is added to 50 cm of 0.100 mol dmNaOH.3)Calculate the pH of the solution formed when 100 cm of 0.050 mol dm HCl is added to 50 cm of 0.500 mol dmKOH.4)Calculate the pH of the solution formed when 10 cm of 1.00 mol dm H2SO4 is added to 25 cm of 1.00 mol dmNaOH.5)Calculate the pH of the solution formed when 50 cm of 0.250 mol dm HNO3 is added to 50 cm of 0.100 mol dmBa(OH)2.6)Calculate the pH change to 100 cm of 0.200 mol dm HCl solution in a flask if 50 cm of 0.100 mol dm NaOH isadded.7)Calculate the pH change to 50 cm of 0.150 mol dm KOH solution in a flask if 50 cm of 0.100 mol dm H2SO4 isadded.3-33-3333-33-33-3-3-33-33-3TASK 5 – a mixture!1)Calculate the pH of the following solutions:-3a) 0.150 mol dm Ba(OH)2-3b) 0.200 mol dm HNO3c)d)-31.500 mol dm H2SO4-30.0500 mol dm NaOH-142-62)a) Calculate the pH of water at 50 C when Kw 5.48 x 10b) is the water neutral? Explain your answer.3)a) 20 cm of 1.0 M H2SO4 with water added to make the volume up to 100 cm .33b) 50 cm of 0.05 M KOH with 200 cm of water added.4)a) Calculate the pH of the solution formed when 100 cm of 0.100 mol dm H2SO4 is added to 50 cm of 0.500-3mol dm NaOH.3-33b) Calculate the pH of the solution formed when 25 cm of 0.250 mol dm HCl is added to 15 cm of 0.100-3mol dm KOH.5)Calculate the pH of the solution formed when 3.5 g of impure sodium hydroxide (98.7 % purity) is dissolved in333water and made up to 100 cm , and then 25 cm of 0.35 mol dm diprotic acid is added.333 www.CHEMSHEETS.co.ukmol dm .14-Jul-12-3Chemsheets A2 00938
WEAK ACIDSStrong acidWeak acidall the molecules break apart to form ionsonly a small fraction of the molecules break apart toform ionsXH–H H X–XX HH–H HXX–X–HX X–H H X -HX H X-HX ¾ H XSome common acids and basesMonoprotic / basicDiprotic / basicStrong acidsWeak acidsStrong basesWeak basesHClhydrochloric acidcarboxylic acidsNaOHsodium hydroxideNH3HNO3nitric acid(e.g. ethanoic acid)KOHpotassium hydroxideH2SO4sulphuric acidBa(OH)2 barium hydroxide The acid dissociation constant (Ka)Ka [H ] [A-][HA]&ammonia-HA ¾ H ApKa -log Ka&Ka 10-pKaThese expressions hold for weakacids at all timesNote-3 Ka – has units mol dm Ka – the bigger the value, the weaker the acid pKa – the smaller the value, the weaker the acidIn a solution of a weak acid in water, with nothing else added: -a) [H ] [A ]b) [HA] [HA]initialKa [H ]2[HA](i.e. the concentration of HA at equilibrium is virtually the same as it was before any of it-3dissociated as so little dissociates, e.g. in a 0.100 mol dm solution of HA, there is virtually-30.100 mol dm of HA)This expression ONLY holds for weak acids in aqueous solution with nothing else added www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 0099
Calculating the pH of a weak acid-3Calculate the pH 0.100 mol dm propanoic acid (pKa 4.87). 2Ka [H ][HA] 2[H ] Ka [HA] [H ] (Ka [HA]) (Ka [HA]) (10-4.87x 0.100]) 1.16 x 10-3-3pH -log (1.16 x 10 ) 2.94-5-3Calculate the concentration of a solution of methanoic acid with pH 4.02 (Ka 1.35 x 10 mol dm ). [H ] 10-4.02 9.55 x 10 2-5-5 2[HA] [H ] (9.55 x 10 )-5Ka1.35 x 10-4-3 6.76 x 10 mol dmTASK 6 – pH of weak acids1)Calculate the pH of the following weak acids:-3a) 0.150 mol dm benenecarboxylic acid (pKa 4.20)-3-5-3b) 0.200 mol dm butanoic acid (Ka 1.51 x 10 mol dm )-3-4-3c) 1.00 mol dm methanoic acid (Ka 1.78 x 10 mol dm )2)Calculate the concentration of the following weak acids.-5-3a) ethanoic acid with pH 4.53 (Ka 1.74 x 10 mol dm )b) pentanoic acid with pH 3.56 (pKa 4.86)3)a) Which is the stronger acid, ethanoic acid (pKa 4.76) or propanoic acid (pKa 4.87)?-53-53b) Which is the stronger acid, propanoic acid (1.35 x 10 mol dm ) or propenoic acid (5.50 x 10 mol dm )?4)Calculate the Ka value for phenylethanoic acid given that a 0.100 mol dm solution has a pH of 2.66.-3Reaction between a weak acid and a strong base--When a weak acid reacts with a strong base, for every mole of OH added, one mole of HA is used up and one mole of A isformed.e.g.HAbefore reaction3after reaction2 lefte.g.HAbefore reaction3after reaction www.CHEMSHEETS.co.uk -OH -A H2O12 made -OH -A H2O107 left3 made14-Jul-12Chemsheets A2 00910
TASK 7 – Reaction of weak acidsWhen the following weak acids react with strong bases the moles of HA left after reaction- the moles of OH left after reaction- the moles of A formed in the reaction1)4 moles of HA with 2.5 moles of NaOH2)6 moles of HA with 1.3 moles of Ba(OH)23)0.15 moles of HA with 0.25 moles of KOH4)0.30 moles of HA with 0.15 moles of NaOH5)100 cm of 0.100 mol dm HA with 50 cm 0.050 mol dm NaOH6)25 cm of 0.500 mol dm HA with 40 cm of 1.0 mol dm KOH7)10 cm of 0.100 mol dm HA with 10 cm of 0.080 mol dm NaOH3-333-333-33-3-3-3Calculating the pH for the solution formed from reaction between a weak acid and a strong base--When a weak acid reacts with a strong base, for every mole of OH added, one mole of HA is used up and one mole of A isformed. 1) Calculate moles HA (it is still HA and not H as it is a weak acid)-2) Calculate moles OH-3) Calculate moles XS HA or OH-If XS HAIf XS OH-4) Calculate [OH ]--5) Use Kw to find [H ]4) Calculate moles HA left and A formed 5) Calculate [HA] leftover and [A ] formed 6) Use Ka to find [H ]6) Find pH7) Find pHNote – if there is XS base, then in terms of working out the pH it is irrelevant whether it was a strong or weak acid as it has allreacted!3-33Calculate the pH of the solution formed when 30 cm of 0.200 mol dm ethanoic acid (pKa 4.76) is added to 100 cm of-30.100 mol dm NaOH.mol HA 30/1000 x 0.200 100/1000 x 0.100-mol OH 0.00600 0.0100-OH is in XS-XS mol OH 0.0100 – 0.00600 0.00400- XS [OH ] 0.00400 0.0308130/1000 XS [H ] Kw 10[OH ] 0.0308 -14-13 3.25 x 10-13pH -log (3.25 x 10 ) 12.49 www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 00911
3-33Calculate the pH of the solution formed when 50 cm of 0.500 mol dm ethanoic acid (pKa 4.76) is added to 75 cm of-30.200 mol dm NaOH.mol HA 50-mol OH /1000 x 0.500 0.025075/1000 x 0.200 0.0150HA is in XSHA before reaction0.0250after reaction0.0100-- OHA H2O0.0150- left over [HA] 0.0100 0.0800125/1000 formed [A ] 0.0150 0.120125/10000.0150- -Ka [H ][A ][HA] [H ] Ka [HA] 10[A ]-4.76-5x 0.0800 1.16 x 100.120-5pH -log (1.16 x 10 ) 4.94Half neutralisation of a weak acid--When half of the HA molecules have reacted with OH , [HA] [A ].3 Ka [H ]orpKa pH-33Calculate the pH of the solution formed when 100 cm of 0.200 mol dm ethanoic acid (pKa 4.76) is added to 40 cm of-30.250 mol dm KOH.mol HA 100-mol OH /1000 x 0.200 0.020040/1000 x 0.250 0.0100HA is in XSHAbefore reaction0.0200after reaction0.0100 - OH-A H2O0.0100-0.0100-half neutralisation and so [HA] [A ] pH pKa 4.76TASK 8 – weak acid strong base3-33-3-4-31)Calculate the pH of the solution formed when 20 cm of 0.100 mol dm methanoic acid (Ka 1.7 x 10 mol dm ) is3-3added to 40 cm of 0.080 mol dm KOH.2)Calculate the pH of the solution formed when 50 cm of 0.500 mol dm propanoic acid (pKa 4.87) is added to 1003-3cm of 0.080 mol dm KOH.3)Calculate the pH of the solution formed when 50 cm of 0.500 mol dm ethanoic acid (pKa 4.76) is added to 503-3cm of 0.250 mol dm KOH.4)Calculate the pH of the solution formed when 50 cm of 0.500 mol dm chloroethanoic acid (pKa 2.86) is added to3-325 cm of 0.100 mol dm Ba(OH)2.5)Calculate the pH of the solution formed when 50 cm of 1.50 mol dm-33-3dm ) is added to 100 cm of 2.00 mol dm KOH.6)Calculate the pH of the solution formed when 25 cm of 1.00 mol dm benzenecarboxylic acid (pKa 4.20) is added3-3to 50 cm of 0.0400 mol dm NaOH.3-33-33 www.CHEMSHEETS.co.uk314-Jul-12-3dichloroethanoic acid (Ka 0.0513 mol-3Chemsheets A2 00912
TASK 9 – Another mixture-31) Calculate the pH of 0.100 mol dm H2SO4.-3-4-32) Calculate the pH of 0.250 mol dm methanoic acid (Ka 1.70 x 10 mol dm )-33) Calculate the pH of 0.20 mol dm Sr(OH)2.3-33-34) Calculate the pH of a mixture of 20 cm of 0.500 mol dm NaOH and 80 cm of 0.200 mol dm HNO3.33-35) Calculate the pH of the solution formed when 100 cm of water is added to 25 cm of 0.100 mol dm NaOH.3-33-36) Calculate the pH of a mixture of 25 cm 0.200 mol dm ethanoic acid (pKa 4.76) and 25 cm 0.100 mol dm NaOH.37) Calculate the pH of a mixture of 100 cm 0.100 mol dmNaOH.3-33ethanoic acid (pKa 4.76) and 50 cm 0.150 mol dm-33-3-38) Calculate the pH of a mixture of 50 cm 0.200 mol dm propanoic acid (pKa 4.87) and 25 cm 1.00 mol dm KOH.pH CURVES & INDICATORS- - Indicators are weak acids where HA and A are different colours. At low pH, HA is the main species present. At high pH, A is the main species present. The pH at which the colour changes varies from one indicator to another. Note that universal indicator is a mixture of indicators and so shows many colours at different pHs. HA ¾ H A-indicatorcolour of HApH range of colour changecolour of A-methyl orangered3.2 - 4.4yellowphenolphthaleincolourless8.2 - 10.0pinkIn a titration, the pH changes rapidly at the end point as the last drop of acid/alkali is added. For an indicator to changecolour at this moment where the moles of acid moles of base, the indicator must change colour within the range of therapid change in pH at the end point. www.CHEMSHEETS.co.uk14-Jul-12Chemsheets A2 00913
pH curves for monoprotic acids (e.g. HCl, HNO3with monoprotic bases-33-3The curves below show the pH as 0.100 mol dm base is added to 25.0 cm of 0.100 mol dm acid:a) strong acid - strong baseb) strong acid - weak basepHpH13137711325cm of base25suitable indicators:3cm of basesuitable indicators:c) weak acid - strong based) weak acid - weak basepHpH13137711cm3 of base2525suitable indicators:Summary:3cm of basesuitable indicators:pHstrong base13weak base7weak acid1strong acid325 www.CHEMSHEETS.co.ukcm of base14-Jul-12Chemsheets A2 00914
TASK 10 – sketching pH curvesSketch each of the following pH curves on the grids shown, and name a suitable indicator.(1)Flask3-33-325 cm 0.10 mol dm HNO3(2)FlaskBurette 50 cm 0.20 mol dm NaOH147700Flaskcm from burette3-33-310 cm 0.20 mol dm HNO3(4)Flaskcm from burette3-33-330 cm 1.00 mol dm NH3Burette 50 cm 1.00 mol dm HClIndicator pH325Burette 50 cm 0.05 mol dm NaOHIndicator pH14147700325Flaskcm from burette3-33-320 cm 0.20 mol dm CH3COOH325(6)Flask3cm from burette-350 cm 0.500 mol dm NH33Burette 50 cm 0.05 mol dm NaOH-3Burette 50 cm 1.00 mol dm methanoic acidIndicator pH-3325(5)3Indicator pH14(3)-3Burette 50 cm 0.10 mol dm HClIndicator pH320 cm 0.10 mol dm NaOHIndicator pH1414770025 www.CHEMSHEETS.co.uk3cm from burette14-Jul-12253cm from buretteChemsheets A2 00915
Titration calculations -H OH H2ORemember these ionic equations 2 H CO3 2- H2O CO2-H HCO3 H2O CO2 H NH3 NH4 TASK 11 – Titration calculations1)33-325.0 cm of a solution of sodium hydroxide required 18.8 cm of 0.0500 mol dm H2SO4.H2SO4 2 NaOH Na2SO4 2 H2O-3a) Find the concentration of the sodium hydroxide solution in mol dm .-3b) Find the concentration of the sodium hydroxide solution in g dm .2)33-325.0 cm of arsenic acid, H3AsO4, required 37.5 cm of 0.100 mol dm sodium hydroxide for neutralisation.3 NaOH(aq) H3AsO4(aq) Na3AsO4(aq) 3 H2O(l)-3a) Find the concentration of the acid in mol dm .-3b) Find the concentration of the acid in g dm .333-33)A 250 cm solution of NaOH was prepared. 25.0 cm of this solution required 28.2 cm of 0.100 mol dm HCl for3neutralisation. Calculate what mass of NaOH was dissolved to make up the original 250 cm solution.4)3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm . 25.0 cm of this solution-33was titrated with 0.095 mol dm NaOH solution, requiring 46.5 cm . Calculate the relative molecular mass of theacid.5)A 1.575 g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250 cm .3One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm of this solution of acid reacted with3-3exactly 15.6 cm of 0.160 mol dm NaOH. Calculate the value of n.6)A solution of a metal carbonate, M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give333-31000 cm of solution. 25.0 cm of this solution reacted with 27.0 cm of 0.100 mol dm hydrochloric acid.Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M.7)A 1.00 g sample of limestone is allowed to react with 100 cm of 0.200 mol dm HCl. The excess acid required3-324.8 cm of 0.100 mol dm NaOH solution. Calculate the percentage of calcium carbonate in the limestone.8)An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm of 0.200 mol d
www.CHEMSHEETS.co.uk 14-Jul-12 Chemsheets A2 009 7 TASK 3 – pH of strong bases 1) Calculate the pH of the following solutions. a) 0.15 mol dm-3 KOH b) 0.05 mol .
www.CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1001 Page 3 As the temperature changes, the average energy of the particles increases and so the
www.CHEMSHEETS.co.uk 10-Mar-2016 Chemsheets A2 1014 Page 7 Example 2 Ethanol has the formula C 2 H 5 OH and is used as a fuel (e.g. for cars in Brazil). It burns .
www.CHEMSHEETS.co.uk 10-Jul-12 Chemsheets AS 029 3 CALORIMETRY The enthalpy change for a reaction can be found by measuring the temperature change in a reaction.File Size: 1MB
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a) [Co(H 2O) 6] 2 Co 2 e) [FeO 4] 2– Fe 6 i) K 2[CoCl 4] Co 2 b) [CrCl 6] 3– Cr 3 f) [Mn(CN) 6] 4– Mn 2 j) K 3[AuF 6] Au 3 c) [Co(NH 3) 6]Cl 2 Co 2 g) [Ni(CO) 4] Ni 0 k) (NH 4) 2[IrCl 6] Ir 4 d) [Co(NH 3) 5Cl]Cl 2 Co 3 h) [Ni(EDTA)] 2– Ni 2 l) Na[Mn(CO) 5] Mn -1 TASK 2 – Writing half equations a) 2I– I 2 2e – b .
metal carbonate metal oxide carbon dioxide (on heating) CaCO 3 CaO CO 2 TASK 4 – WRITING BALANCED EQUATIONS 1) Balance the following equations.
Creating an icon button This example will be using the class md-icon-button, which must be applied to in order to get an icon button. It is also recommended to add an aria-label attribute to for accessibility purpose or the ARIA provider will throw a warning that there is no aria-label. Usually, there is an element in the attribute.
