Module 8 Unit 2 Assignment B

MODULE 8 UNIT 2 Assignment Part B Solve the following problems. Show all calculations. 1. A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? 4 marks 2HCl (aq) + Ba(OH) 2 (aq)  BaCl 2 (aq) + 2H 2 O (l) Mol 2HCl = 0.025L x 0.135 mol/L = 0.003375 or 3.375 x 10 -3 Molar ratios: 2mol HCl / 1mol Ba(OH) 2 Moles of acid: 3.375 x 10 -3 x 1mol Ba(OH) 2 / 2mol HCl = 0.0016875 or 1.6875 x 10 -3 The concentration of the barium hydroxide solution is 0.072 M Ba(OH) 2 2. A flask contains 41.04 mL of a solution of potassium hydroxide. The solution is titrated and reaches an equivalence point when 21.65 mL of a 0.6515 M solution of HNO 3 is added. Calculate the molarity of the base solution. 4 marks KOH (aq) + HNO 3 (aq)  KNO 3 (aq) + H 2 O (l) Molar mass of HNO 3 = 63.01g Moles of acid = 0.02165L HNO 3 x 1mol / 63.01g = 0.0003436 or 3.436 x 10 -4 mol KNO 3 3.436 x 10 -4 mol KNO 3 x 1mol KNO 3 / 1mol KOH = 3.436 x 10 -4 mol KOH M KOH = 3.436 x 10 -4 KOH / 0.04104L = 0.00837 M KOH 3. What is the molar mass of an acid if a 0.363 g sample of the triprotic acid (H 3 A) requires 23.44 mL of a 0.135 M NaOH solution? (H 3 A is a generic designation for an unknown acid with 3 H + ions and the A 3- ion is a generic designation for our unknown anion.) NaOH + H 3 A  NaN 3- + H 2 O Mol NaOH= 0.135m x 0.02344L = 0.0031644 or 3.1644 x 10 -3 1mol NaOH = 1 mol H 3 A = 3.1644 x 10 -3 Molar mass= 0.363g / 3.1644 x 10 -3 = 114.713g H3 A The molar mass of the acid used in this titration is 114.713g H 3 A Submission Instructions 1