Titrations 1 Past Paper PDF

TITRATIONS 1 3 3 3 1 25.0 cm of 0.200 mol/dm barium hydroxide solution reacted with 22.8 cm of hydrochloric acid. Calculate th...

TITRATIONS 1 3 3 3 1 25.0 cm of 0.200 mol/dm barium hydroxide solution reacted with 22.8 cm of hydrochloric acid. Calculate the 3 concentration of the hydrochloric acid in mol/dm. Give your answer to 3 significant figures. Ba(OH)2(aq) + 2 HCl(aq) → BaCl2(aq) + 2 H2O(l) 3 𝟐𝟓.𝟎 moles Ba(OH)2 = conc x vol (dm ) = 0.200 x = 0.00500 mol 𝟏𝟎𝟎𝟎 moles HCl = 2 x moles of Ba(OH)2 = 2 x 0.00500 = 0.0100 mol 𝐦𝐨𝐥𝐞𝐬 𝟎.𝟎𝟏𝟎𝟎 3 conc HCl = = 𝟐𝟐.𝟖 = 0.439 mol/dm 𝐯𝐨𝐥𝐮𝐦𝐞 (𝐝𝐦𝟑 ) 𝟏𝟎𝟎𝟎 3 3 3 2 22.5 cm of sodium hydroxide solution reacted with 25.0 cm of 0.100 mol/dm hydrochloric acid. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 3 a) Calculate the concentration of the sodium hydroxide solution in mol/dm. Give your answer to 3 significant figures. 3 𝟐𝟓.𝟎 moles HCl = conc x vol (dm ) = 0.100 x = 0.00250 mol 𝟏𝟎𝟎𝟎 moles NaOH = moles of HCl = 0.00250 mol 𝐦𝐨𝐥𝐞𝐬 𝟎.𝟎𝟎𝟐𝟓𝟎 3 conc NaOH = = 𝟐𝟐.𝟓 = 0.111 mol/dm 𝐯𝐨𝐥𝐮𝐦𝐞 (𝐝𝐦𝟑 ) 𝟏𝟎𝟎𝟎 3 b) Calculate the concentration of the sodium hydroxide solution in g/dm. Give your answer to 3 significant figures. 3 conc NaOH = 40 x 0.111 = 4.44 g/dm 3 3 3 3 What volume of 0.150 mol/dm rubidium hydroxide reacts with 25.0 cm of 0.240 mol/dm nitric acid? Give your answer to 3 significant figures. RbOH(aq) + HNO3(aq) → RbNO3(aq) + H2O(l) 3 𝟐𝟓.𝟎 moles HNO3 = conc x vol (dm ) = 0.240 x = 0.00600 mol 𝟏𝟎𝟎𝟎 moles RbOH = moles HNO3 = 0.00600 mol 𝐦𝐨𝐥𝐞𝐬 𝟎.𝟎𝟎𝟔𝟎𝟎 3 volume RbOH = = = 0.0400 dm 𝐜𝐨𝐧𝐜 𝟎.𝟏𝟓𝟎 3 3 3 4 25.0 cm of 0.200 mol/dm sodium hydroxide solution reacted with 28.7 cm sulfuric acid. Calculate the concentration 3 of the sulfuric acid in mol/dm. Give your answer to 3 significant figures. 2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) 3 𝟐𝟓.𝟎 moles NaOH = conc x vol (dm ) = 0.200 x = 0.00500 mol 𝟏𝟎𝟎𝟎 𝟏 𝟏 moles H2SO4 = x moles of NaOH = x 0.00500 = 0.00250 mol 𝟐 𝟐 𝐦𝐨𝐥𝐞𝐬 𝟎.𝟎𝟎𝟐𝟓𝟎 3 conc H2SO4 = = 𝟐𝟖.𝟕 = 0.0871 mol/dm 𝐯𝐨𝐥𝐮𝐦𝐞 (𝐝𝐦𝟑 ) 𝟏𝟎𝟎𝟎 © www.CHEMSHEETS.co.uk 22-May-2018 Chemsheets GCSE 1105  3 3 3 5 25.0 cm of 0.150 mol/dm sodium hydroxide reacted with 30.3 cm of a solution of ethanoic acid. CH3COOH(aq) + NaOH(aq) → CH3COONa (aq) + H2O(l) 3 a) Calculate the concentration of the ethanoic acid in mol/dm. Give your answer to 3 significant figures. 3 𝟐𝟓.𝟎 moles NaOH = conc x vol (dm ) = 0.150 x = 0.00375 mol 𝟏𝟎𝟎𝟎 moles CH3COOH = moles NaOH = 0.00375 mol 𝐦𝐨𝐥𝐞𝐬 𝟎.𝟎𝟎𝟑𝟕𝟓 3 conc CH3COOH = = 𝟑𝟎.𝟑 = 0.124 mol/dm 𝐯𝐨𝐥𝐮𝐦𝐞 (𝐝𝐦𝟑 ) 𝟏𝟎𝟎𝟎 3 b) Calculate the concentration of the ethanoic acid in g/dm. Give your answer to 3 significant figures. 3 conc CH3COOH = 60 x 0.124 = 7.43 g/dm Area Strength To develop Area Strength To develop Area Strength To develop Done with care and thoroughness Uses equation for other reactant moles Does not round too much Shows suitable working Can find concentration of other reagent Can use sig figs Can work out moles from conc & vol Can convert mol/dm to g/dm 3 3 Gives units © www.CHEMSHEETS.co.uk 22-May-2018 Chemsheets GCSE 1105