24 Buffers - 33.3 mL of 0.2 M HCl is added to For E and EH+...

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1 1 33.3 mL of 0.2 M HCl is added to 25.0 mL of a 0.3 M solution of ephedrine . What is the pH of the resulting solution? A) 8.31 B) 6.06 C) 7.00 D) 9.06 E) 9.96 For E and EH + K b = 9.1x10 -5 K a = 1.1x10 -10 9.96 was the pH before the acid was added, note how little the pH changed! Why? 1. Determine millimoles of E present initially. 2. Determine millimoles of HCl added. 3. Determine millimoles of EH + produced. 4. Determine millimoles of E remaining. 5. Use K a to calculate the hydronium ion concentration and hence the pH. The base E consumed the strong acid to make a weak acid! 2 Construct a RICE table for HCl reacting with E. Initial amount of HCl and E: HCl: (0.2 M x 33.3 mL) = 6.66 mmoles E: (0.3 M x 25.0 mL) = 7.50 mmoles HCl + E EH + + Cl - 6.66 7.50 0 0 mmoles -6.66 -6.66 +6.66 +6.66 0 0.84 +6.66 +6.66 Major species: H 2 O, E, EH + , Cl Equilibria involving E, EH + , and H 2 O determine the pH. Chloride is a very very weak base and isn’t involved. It is the conjugate base of a strong acid. Strong acid reacting with a base. Reaction goes to completion.
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2 3 Since the pH is requested, work with the acid ionization equilibrium. EH + + H 2 O E + H 3 O + ] [EH ][E] O [H K 3 a In terms of concentration, volume = 25.0 mL + 33.3 mL 6.66/58.3 mL 0.84/58.3 mL 0 -x x x 6.66/58.3-x 0.84/58.3+x x HCl + E EH + + Cl 6.66 7.50 0 0 mmoles -6.66 -6.66 +6.66 +6.66 0 0.84 +6.66 +6.66 4 Equilibrium concs in M EH + + H 2 O E + H 3 O + 6.66/58.3-x 0.84/58.3+x x For E and EH + K b = 9.1x10 -5 K a = 1.1x10 -10 x) - (6.66/58.3 x) 8.3 (x)(0.84/5 10 1.1 ] [EH ][E] O [H K 10 - 3 a x 9.06 pH 0.84 6.66 10 1.1 x 6.66 0.84 x 10 1.1 10 10 x x 25.0 mL + 33.3 mL = 58.3 mL Shortcut!
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3 5 The Buffer Region in an Acid – Base Titration The region where both a weak acid and a conjugate base are major species in the solution.
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24 Buffers - 33.3 mL of 0.2 M HCl is added to For E and EH+...

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