WS5_key - Chem 142 Name Summer 2011 Section Worksheet 5...

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Chem 142 Summer 2011 Page 1 of 7 Name ________________________________________ Section ________________ Worksheet 5: Reactions in Solution – Acid/Base and Redox 1. How many mL of a 0.800 M NaOH solution is needed to neutralize 40.00 mL of a 0.600 M HCl solution? Molecular Rxn: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Net Ionic Rxn: H + (aq) + OH - (aq) H 2 O (l) moles H + = 0.04000 L ! 0.600 mol HCl 1 L ! 1 mol H + 1 mol HCl = 0.0240 mol H + moles OH ! = 0.0240 mol H + " 1 mol OH ! 1 mol H + " 1 mol NaOH 1 mol OH ! = 0.0240 mol NaOH volume NaOH = 0.0240 mol NaOH ! 1 L 0.800 mol NaOH = 0.0300 L NaOH solution volume NaOH = 0.0300 L NaOH ! 1000 mL 1 L = 30.0 mL NaOH solution 2. What volume of each of the following acids will react completely with 50.00 mL of 0.100 M NaOH? moles OH ! = 0.05000 L solution " 0.100 mol NaOH 1 L solution " 1 mol OH ! 1 mol H + " 1 mol OH ! 1 mol NaOH = 0.00500 mol OH ! a. 0.100 M HCl moles H + = 0.00500 mol OH ! " 1 mol H + 1 mol OH ! " 1 mol HCl 1 mol H + = 0.00500 mol HCl volume HCl = 0.00500 mol HCl ! 1 L 0.100 mol HCl = 0.0500 L HCl solution b. 0.100 M H 2 SO 3 (two acidic hydrogens) moles H + = 0.00500 mol OH ! " 1 mol H + 1 mol OH ! " 1 mol H 2 SO 3 2 mol H + = 0.00250 mol H 2 SO 3 volume H 2 SO 3 = 0.00250 mol H 2 SO 3 ! 1 L 0.100 mol H 2 SO 3 = 0.0250 L H 2 SO 3 solution
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Chem 142 Summer 2011 Page 2 of 7 c. 0.200 M H 3 PO 4 (three acidic hydrogens) moles H + = 0.00500 mol OH ! " 1 mol H + 1 mol OH ! " 1 mol H 3 PO 4 3 mol H + = 0.00167 mol H 3 PO 4 volume H 3 PO 4 = 0.00167 mol H 3 PO 4 ! 1 L 0.200 mol H 3 PO 4 = 0.00835 L H 3 PO 4 solution d. 0.150 M HNO 3 moles H + = 0.00500 mol OH ! " 1 mol H + 1 mol OH ! " 1 mol HNO 3 1 mol H + = 0.00500 mol HNO 3 volume HNO 3 = 0.00500 mol HNO 3 ! 1 L 0.150 mol HNO 3 = 0.0333 L HNO 3 solution e. 0.200 M HC 2 H 3 O 2 (one acidic hydrogen) moles H + = 0.00500 mol OH ! " 1 mol H + 1 mol OH ! " 1 mol HC 2 H 3 O 2 1 mol H + = 0.00500 mol HC 2 H 3 O 2 volume HC 2 H 3 O 2 = 0.00500 mol HC 2 H 3 O 2 ! 1 L 0.200 mol HC 2 H 3 O 2 = 0.0250 L HC 2 H 3 O 2 solution f. 0.300 M H 2 SO 4 (two acidic hydrogens) moles H + = 0.00500 mol OH ! " 1 mol H + 1 mol OH ! " 1 mol H 2 SO 4 2 mol H + = 0.00250 mol H 2 SO 4 volume H 2 SO 4 = 0.00250 mol H 2 SO 4 ! 1 L 0.300 mol H 2 SO 4 = 0.00833 L H 2 SO 4 solution 3. A 25.00 mL solution of sodium hydroxide is titrated to the endpoint (the point where the acid and base have reacted to be completely neutralized) after the addition of 42.95 mL of a 0.1067 M HCl solution.
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WS5_key - Chem 142 Name Summer 2011 Section Worksheet 5...

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