University:
Glendale Community College
Course:
CHM130 | Fundamental Chemistry
Academic year:

2019
Views:

354

Pages:

2
Author:

Simeon Barrera

Buffer and Titration Key 1. Write the molecular, ionic and net ionic reactions for adding hypochlorous acid (HClO) with calcium hydroxide. molecular__ 2 HClO(aq) + Ca(OH)2 (aq) J 2 H2O(l) + Ca(ClO)2(aq) ionic_ net ionic_ 2 HClO(aq) + Ca2+(aq) + 2 OH-(aq) HClO(aq) + OH-(aq) J 2 H2O(l) + Ca2+(aq) + 2 ClO-(aq) J H2O(l) + ClO-(aq) 2. Calculate the pH of a 2.00 L solution containing 0.885 moles of hypochlorous acid (HClO) and 0.905 moles of NaClO. Given Ka for HClO is 3.0 x 10-8. This is a buffer. We can use the buffer equation. pKa = 7.5229 ⎡ 0.905mol / 2.00 L ⎤ = 7.53 pH = 7.5229 + log ⎢ ⎣ 0.885mol / 2.00 L ⎥⎦ 3. Calculate the pH if 50.0 mL of 0.125M nitric acid is added to a 2.00L buffer system composed of 0.250M acetic acid and 0.250M lithium acetate. Acetic acid moles = acetate ion moles = 2.00L ( 0.250 mol / L) = 0.500 moles Nitric acid moles = 0.0500L (0.125 mol / L) = 0.00625 mol Ka from text = 1.8 x 10-5 so pKa = 4.7447 Rxn is the H+ from nitric acid with the c. base of the buffer or acetate ion Rxn H+ (aq) + CH3COO- (aq) J HCH3COO-(aq) I 0.00625 mol 0.500mol 0.500 mol C - 0.00625 - 0.00625 + 0.00625 Final 0 0.49375 mol 0.50625 mol All the nitric acid is used up and reacted, and we still have buffer left, use buffer equation ⎡ 0.49375mol / 2.05L ⎤ = 4.73 pH = 4.7447 + log ⎢ ⎣ 0.50625mol / 2.05L ⎥⎦ 4. How many milliliters of 0.95M NaOH must be added to 35.0 mL of 0.85M CH3COOH to reach the equivalence point? What is the pH at the equivalence point? Given: Ka for acetic acid is 1.8 x 10-5 Finding the volume of NaOH is stoichiometry. 35.0 mL ( L / 1000 mL)(.85 mol / L)( 1 NaOH / 1 acid)( L / 0.95 mol)( 1000mL / L) = 31 mL To find the pH we need the reaction with IF table in moles. So calc all the moles and put in table: Rxn I C F CH3COOH(aq) + NaOH(aq) 0.02975 mol .02975 mol - 0.02975 - 0.02975 0 0 J NaCH3COO (aq) + H2O (l) 0 ---+ 0.02975 ----0.02975 mol ---- This makes sense because at the equivalence point all the acid and base are used up and only salt remains. This salt, sodium acetate, is a basic salt. The sodium ion is neutral, but the acetate ion is basic. This is an ICE table so we need molarity in it, not moles. Basically now we find the pH of a salt solution. Rxn I C E CH3COO- (aq) + H2O (l) ' CH3COOH (aq) + OH- (aq) .4508 M -0 0 -x -x x .4508 – x -x x Kb = Kw / Ka = 5.5556 x 10-10 = x2 / .4508 X = 1.58 x 10-5 approximation checks pH = 9.20 5. Calculate the pH when 25.0 mL of 0.100M HBr is added to 15.0 mL of 0.100M LiOH. Rxn I C F HBr(aq) + LiOH (aq) J LiBr(aq) + H2O(l) .00250 mol .00150 mol 0 --.00150 -.00150 +.00150 -.00100 mol 0 .00150 mol -- [H+] = .00100 mol / .0400 L = .0250 M so pH = 1.602