PPSet_2_AnsKey - Tufts University Chem 2 Summer 2008 Paper...

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Tufts University – Chem 2 – Summer 2008 Paper Problem Set 2 (7 problems, 55 points total) Answer Key 1 . (10 pts) A buffer solution is prepared by mixing 500. mL of 0.220 M C 6 H 5 COOH(aq) with 500. mL of 0.100 M NaOH(aq). Estimate the osmotic pressure (in atm) of the buffer solution at 20 o C assuming that the reaction goes to completion, strong electrolytes dissociate in full, and the dissociation of weak electrolytes is negligible. Acid-base reaction: C 6 H 5 COOH + NaOH Na(C 6 H 5 COO) + H 2 O. Initial number of moles: 0.500 L × 0.220 M = 0.110 mol C 6 H 5 COOH, 0.500 L × 0.100 M = 0.0500 mol NaOH. The reaction is 1:1; thus, NaOH is limiting. 0.110 mol C 6 H 5 COOH - 0.0500 mol NaOH = 0.060 mol C 6 H 5 COOH excess. Benzoic acid (like all organic acid) is a weak electrolyte and will stay largely molecular in solution. 0.0500 mol Na(C 6 H 5 CO 2 ) produced. Sodium benzoate is a salt; it is a strong electrolyte and dissociates is full: Na(C 6 H 5 COO)(aq) Na + (aq) + C 6 H 5 COO - (aq); 0.0500 mol Na + and 0.0500 mol C 6 H 5 COO - produced in the solution. 0.060 mol C 6 H 5 COOH(aq) is the leftover. Total V after mixing= 1.000 L. Total molarity of all solute ions and molecules in the solution after reaction = 0.0500 M Na + (aq) + 0.0500 M C 6 H 5 COO - (aq) + 0.060 M C 6 H 5 COOH = 0.160 M.
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PPSet_2_AnsKey - Tufts University Chem 2 Summer 2008 Paper...

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