# Ch6 Example Problems Answers - HODSON CHE 1301 Petrucci Ch5...

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HODSON CHE 1301 Petrucci Ch5 Example Problems The following problems are taken from old exams . ..... 1) What is the concentration of Br (aq) in a solution prepared by mixing 75.0 mL of 0.62 M iron(III) bromide with 75.0 mL of water? Assume volumes are additive. a) 0.31 M b) 1.9 M c) 0.93 M d) 0.62 M Answer: C Explanation: We have to find the molarity (moles/L) of bromide ion (Br ) in this solution. We initially need to find the number of moles of iron (III) bromide (FeBr 3 ) we’re adding to the water (the molaritry of the solution gives us the equivalence 0.62 moles = 1 L, and from this we write the appropriate conversion factor), and from this we can find the moles of Br (because looking at the formula, there are 3 Br ions in each FeBr 3 unit). The total volume of solution wil be 75.0 mL + 75.0 mL = 150.0 mL = 0.1500 L. So…. . Moles FeBr 3 = 75.0 mL × (1 L / 1000 mL) × (0.62 moles FeBr 3 / 1 L) = 0.0465 moles FeBr 3 The next step involves recognizing that when we dissolve the ionic compound FeBr 3 in water, it splits into an Fe 3+ (aq) ion and three Br (aq) ions (we put the (aq) symbol after these to indicate that they’re dissolved in water). 0.0465 moles FeBr 3 × (3 moles Br / 1 mole FeBr 3 ) = 0.1395 moles Br ions These are present in the total volume of solution = 0.1500 L, therefore molarity is . . 0.1395 moles Br ions / 0.1500 L = 0.93 M 2) Which of the following pairs of aqueous solutions will give a precipitate when mixed? a) K 3 PO 4 and CaCl 2 b) AgNO 3 and Ca(ClO 4 ) 2 c) NaCl and Hg(CH 3 COO) 2 c) LiClO 4 and Ba(OH) 2 Answer: A Explanation: This is just testing the solubility rules. Either look to the notes (slide 17) or to the back of the quickhits sheet. We look at the cations and anions in these compounds, and ‘swap partners’ to give the alternate cation / anion pairs. In solution all the ions will be free to move around, and we’re essentially wondering whether they’ll bump into each other and form a precipitate (insoluble compound). So, switch and swap the ions in each answer and see if the resulting compounds would be insoluble according to our rules. a) K 3 PO 4 and CaCl 2 gives Ca 3 (PO 4 ) 2 and KCl Going by our rules, rule one says Grp 1A salts (ionic compounds) are soluble, so we expect the second compound as written to dissolve up. Also, Chloride is soluble by rule 4. Phosphate is insoluble by rule 5, so we would expect the first compound written above to give a precipitate.

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b) AgNO 3 and Ca(ClO 4 ) 2 gives Ag(ClO 4 ) and Ca(NO 3 ) 2 Percholrates (ClO 4 ) are soluble by rule 2, so the first compound is expected to be soluble. Nitrates are also expected to be soluble by rule 2, so the second compound is also expected to be soluble. The answer was a), so you get the picture…… switch and swap the cation/anion pairs
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## This note was uploaded on 12/18/2009 for the course CHE 1301 taught by Professor Klausmeyer during the Spring '08 term at Baylor.

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Ch6 Example Problems Answers - HODSON CHE 1301 Petrucci Ch5...

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