1aa3-2012-tut2-a

# 1aa3-2012-tut2-a - CHEMISTRY 1AA3 Week of JANUARY 16, 2012...

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Unformatted text preview: CHEMISTRY 1AA3 Week of JANUARY 16, 2012 TUTORIAL PROBLEM SET 2 Solutions ______________________________________________________________________________ Page 1 of 7 1. Which of the following combination(s) of reagents will result in a buffer? a. 100 mL 0.1M NaOH plus 50 mL 0.1M CH 3 COOH b. 100 mL 0.1M NaOH plus 100 mL 0.1M CH 3 COOH c. 100 mL 0.1M KOH plus 100 mL 0.2M CH 3 COOH d. 100 mL 0.1M NH 3 plus 50 mL 0.2M HCl(aq) e. 100 mL 0.1M NH 3 plus 50 mL 0.1M HBr(aq) Solution: a. 100 mL 0.1M NaOH plus 50 mL 0.1M CH 3 COOH not a buffer, but a combination of weak and strong base b. 100 mL 0.1M NaOH plus 100 mL 0.1M CH 3 COOH not a buffer, but a weak base (NaCH 3 COO- ) c. 100 mL 0.1M KOH plus 100 mL 0.2M CH 3 COOH BUFFER! d. 100 mL 0.1M NH 3 plus 50 mL 0.2M HCl(aq) not a buffer, but a slightly acidic salt e. 100 mL 0.1M NH 3 plus 50 mL 0.1M HBr(aq) BUFFER! 2. Describe the pH of a buffer solution consisting of 100 mL containing 0.1M CH 3 COOH and 0.125M CH3COO- , both before and after addition of 10 mL of 0.25M HCl. K a = 1.8 x 10-5 . The Henderson-Hasselbalch equation pH = pKa + log([A-]/[HA]) can be used to calculate both pH values: i. Before addition of acid: pH = -log(1.8 x 10-5) + log(0.125/0.1) = 4.74 + 0.1 = 4.74 ii. After addition of acid: Addition of 10ml 0.25M HCl (0.0025moles) will convert the equivalent amount of acetate anion into acetic acid. This will lead to a new ratio of moles acetate to moles acetic acid of 0.0125 moles acetic acid to 0.01 moles acetate, in 110 mL new total volume. Here two options arise: a. An ICE table could be used to calculate the pH from equilibration of the resulting new concentrations of acetate and acetic acid. b. Alternatively, the Henderson-Hasselbalch equation could be used directly. As the HH eq. Only considers rations of the two buffer components, but not their concentrations, you simply enter the new moles of acid (acetic acid) and base (acetate) into the HH eq, and obtain: pH = pKa + log(0.01/0.0125) = 4.74 -0.1 = 4.64 2 3. Which of the three acids given below would be appropriate to prepare a buffer solution to maintain the pH at 4.30? What ratio of the acid/conjugate base should be used to fix the pH at 4.30? pH at 4....
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1aa3-2012-tut2-a - CHEMISTRY 1AA3 Week of JANUARY 16, 2012...

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