# ch2sol - CH201 SI: CHAPTER 2 PRACTICE TEST

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Unformatted text preview: CH201 SI: CHAPTER 2 PRACTICE TEST  1. Nitric acid (HNO3, Mm = 63.02 g/mol) is a strong acid that fully dissociates into  H1+ and NO31‐ in water.  A 2.000% by volume aqueous solution of Nitric Acid (HNO3,  Mm = 63.02 g/mol) has a density of 1.012 g/mL.  What is the molarity of NO31‐ ions?  2.000% by vol. HNO3 = 2.000 mL HNO3 / 100.0 mL Solution  100.0 mL Solution * 1.012g*mL­1 = 101.2 g Solution  If the solution is 2 % by volume HNO3, then 100.0 mL solution contains 98.00 mL water  98.00 mL water * 1.0 g*mL water (density of water) = 98.00 g water  So, 100mL of solution has a mass of 101.2 g, 98.00 g of which is water.  101.2g solution – 98.00 g water = 3.2 g HNO3 /63.02 g*mol­1 HNO3 = 0.051 mol HNO3  0.051 mol HNO3 * (1 mol NO31­/1 mol HNO3) = 0.051 mol NO31­/ 0.100 L = 0.51 M   [NO31­] = 0.51 M  2.  If 30mL of the solution from #1 requires 17.32 mL of a NaOH solution to  completely neutralize, what is the molarity of OH1‐ in the NaOH solution (NaOH  completely dissociates into Na1+ and OH1‐ ions in water).  Neutralization Reaction:    H1+ + OH1‐  H2O   0.51 mol*L­1 NO31­ * 0.030 L = 0.0153 mol NO31­ * (1 mol H1+/1 mol NO31­) =   0.0153 mol H1+ * (1 mol OH1­/1 mol H1+) = 0.0153 mol OH1­ / 0.01732 = 0.8834 M   [OH1­] = 0.8834 M  3. What mass of Ca(OH)2 is required to make I.00 L of solution with the same  concentration of OH1‐ as #2?   0.8834 mol*L­1 OH1­ * 1.00 L = 0.8834 mol OH1­ * (1 mol Ca(OH)2/2 mol OH1­) =   0.4416 mol Ca(OH)2 * 74.08 g*mol­1 Ca(OH)2 = 32.71 g Ca(OH)2  32.71g Ca(OH)2  4. What is the mole fraction of OH1‐ ions in the solution from #3?  Choose 1.00L solution to make the calculations easier.    0.4416 mol Ca(OH)2* (1 mol Ca2+/1 mol Ca(OH)2) = 0.4416 mol Ca2+  0.4416 mol Ca(OH)2*(2 mol OH­1/1 mol Ca(OH)2) = 0.8834 mol OH1­  1000. mL H20* 1 g*mL­1 H2O = 1000. g HzO / 18.00 g*mol­1 H2O = 55.55 mol H2O  XOH1­ = mol OH1­/ total mol solution = mol OH1­/ (mol Hz0 + mol of Ca2+ + mol H20)  = 0.8834 / (55.55 + 0.4416 + 0.8834) = 0.0155    XOH1­ = 0.0155  ...
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## This note was uploaded on 01/27/2011 for the course CHEM 201 taught by Professor Wilson during the Spring '10 term at N.C. Central.

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