Analytical+Chemistry (1)

# Analytical+Chemistry (1) - worksheet.. problem solving

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Analytical Chemistry Solutions to Workshop Problems due Week 2 1. ( a) Calculate the p-function of the ion-product constant of water, K w , at 25C. (b) Calculate the p-function of the concentrations of the four ions in a solution that is simultaneously 2.00 x 10 -3 M in NaCl and 5.4 x 10 -4 M in HCl. [Look at part (a) to get a hint as to the nature of the fourth ion.] (a) The value of the ion product of water, K w is given on page 101 and a Table 6-1 shows its temperature dependence. The value is 1.01 x 10 -14 at 25C pK w = - log (1.01 x 10 -14 ) = 13.996 (a value of 14.0 or 14.00 is fine) (b) The concentration of H + is equal to the concentration of HCl since one H + is present in solution for each HCl molecule that was dissolved in the solution. The same is true of the Na + cation. It’s concentration is equal to the concentration of the NaCl in the solutions since one Na + is present fro each NaCl molecule that dissolved. So pH = - log [H + ] = - log (5.4 x 10 -4 ) = -(-3.2676) = 3.27 pNa = - log (2.00 x 10 -3 ) = 2.699 Notice that these answers have different significant figures as the given concentrations have different sig figs. Cloride anions in the solution come from both the HCl and NaCl so the total Cl - concentration is given by the sum of the concentrations of the two solutes: [Cl - ] = 2.00 x 10 -3 M + 5.4 x 10 -4 M = 2.54 x 10 -3 M pC1 = - (log 2.54 x 10 -3 ) = 2.595 Finally the other anion present is the hydroxide ion. If you know the concentration of H + in a solution you can use K w to calculate the hydroxide ion concentration. K w = [H + ][OH - ] solving for hydroxide gives [OH - ] = K w / [H + ] [OH - ] = K w / [H + ] = 1.00 x 10 -14 / 5.4 x 10 -4 = 1.852 x 10 -11 pOH = - log (1.852 x 10 -11 ) = 10.73 (has 2 sig figs since 5.4 has two sig figs) If you understand logs well you remember that if dividing numbers is the same as subtracting their logs. So a quick way to get pOH if pH is know is the following pOH = 14.00 – pH = 14.00 - 3.27 = 10.73 1

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2. Magnetite is a mineral having the formula Fe 3 O 4 or FeO Fe 2 O 3 . A 1.1324-g sample of a magnetite ore was dissolved in concentrated HCl to give a solution that contained a mixture of Fe 2+ and Fe 3+ ions. Nitric acid was added and the solution was boiled for a few minutes, which converted all of the iron ions to Fe 3+ . The Fe 3+ was then precipitated as Fe 2 O 3 xH 2 O by addition of NH 3 . {The x here refers to a variable amount of water being included in the solid precipitate.] After filtration and washing, the residue was ignited at a high temperature to give 0.5394 g of pure Fe 2 O 3 (159.69 g/mol). [Heating drives off the water so that a stoichiometric solid with the formula Fe 2 O 3 is left behind.] Calculate (a) the percent by weight of Fe (55.847 g/mol) in the original ore sample and (b) the percent Fe 3 O 4 (231.54 g/mol) in the original ore sample. Like most stoichiometry problems you start with the known information (usually given at the end of
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## This note was uploaded on 01/12/2011 for the course CHEM 101 taught by Professor Mrs.taneji during the Fall '10 term at Abant İzzet Baysal University.

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Analytical+Chemistry (1) - worksheet.. problem solving

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