worksheet_06_103 (1)

1 CHEMISTRY 103 – WORKSHEET #6 – Module 4 Stoichiometry (Part III) Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) MOLARITY: concentration of solutions; abbreviated as M ( “molar”) Solute : chemical that dissolves; Solvent : liquid solute dissolves in; Solution : Solvent + Solute 1. Molarity calculations (ID this type of problem: 1 chemical; 1 concentration) : use: Example 1: How many grams Na 2 CO 3 are needed to prepare 650ml of a 2.25M Na 2 CO 3 solution? Answer 1: 155g Na 2 CO 3 {1 chemical and 1 concentration ® use ; ; x = 1.4625mol Na 2 CO 3 ; } 2. Dilution Problems (ID this type of problem: 1 chemical; 2 concentrations) : use: M 1 V 1 = M 2 V 2 equation used only with dilutions (1 = initial, 2 = final; V 2 = total volume = V 1 + water) Example 2: 2.00ml of 2.5 x 10 -4 M Zn(NO 3 ) 2 (aq) was diluted in a volumetric flask to 25.00ml. 5.00ml of this new solution was removed with an Eppendorf pipette and diluted again in a new volumetric flask to 25.00ml. What is the final concentration of the Zn(NO 3 ) 2 ? Answer 2: 4.0 x 10 -6 M (1 chemical and 2 concentrations ® a dilution; use M 1 V 1 = M 2 V 2 two times); M 1 = 2.5 x 10 -4 M; V 1 = 2.00ml; M 2 = x; V 2 = 25.00ml; (2.5 x 10 -4 M)(2.00) = M 2 (25.00); M 2 = 2.0 x 10 -5 M; use M 1 V 1 = M 2 V 2 again: M 1 = 2.0 x 10 -5 M; V 1 = 5.00ml; M 2 = x; V 2 = 25.00ml; (2.0 x 10 -5 M)(5.00) = M 2 (25.00); M 2 = 4.0 x 10 -6 M; Note: you can use ml in the dilution equation as long as both volumes are in ml 3. Stoichiometry (ID this type of problem: 2 chemicals reacting) : can be referred to as neutralization, titration, completely reacts – both reactants run out; use flowchart Example 3: If it required 56.0ml of a 0.250M H 2 SO 4 solution to neutralize 26.5ml NaOH, what was the original concentration of the NaOH? Answer 3: 1.06M NaOH (2 chemicals ® stoichiometric problem; M A ® M B question; 3 steps from the above flow chart) Step 1: find mol H 2 SO 4 : M = mol/L; Step 2: convert mol H 2 SO 4 to mol NaOH using a balanced reaction; [hint: the stoichiometric ratio between an acid and base can be determined by inspection without writing the reaction; the ratio must be: 1 H + to 1 OH - or in other words, the same number of H + and OH - ; since H 2 SO 4 has 2 H + and NaOH has 1 OH - , there needs to be 1 H 2 SO 4 (2 H + ) and 2 NaOH (2 OH - ): H 2 SO 4 (aq) + 2NaOH(aq) ® 2H 2 O(l) + Na 2 SO 4 (aq) Step 3: calculate the concentration using M NaOH = mol NaOH /L NaOH ; M = mol solute L solution M = mol L 2.25M = x mol (650ml)(1L /1000ml) 1 . 4625 mol Na 2 CO 3 106 . 0 g Na 2 CO 3 1 mol Na 2 CO 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 155 . 03 g Na 2 CO 3 atoms or molecules B atoms or molecules A grams A grams B moles A moles B 1 mol B = 6.022 x 10 B 23 1 mol A = 6.022 x 10 A 23 molar mass A molar mass B molarity A molarity B M = mol /L A A A M = mol /L B B B chemical formula or chemical reaction mol H 2 SO 4 = M x L = 0.250M (56.0ml)(1L) 1000ml ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.0140mol H 2 SO 4 0 . 0140 mol H 2 SO 4 2 mol NaOH 1 mol H 2 SO 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0 . 0280 mol NaOH M NaOH = 0.0280mol NaOH (26.5ml)(1L /1000ml) = 1.057M