105 HE1 F96

Gravimetric Analysis, Volumetric Analysis, Statistics and Ksp

Tuesday, October 1, 1996 Afternoon

100 points (exam sums to 100) Name _________________________________.

Directions: Answer all questions in the spaces provided. Solutions must show complete set-ups, including units, for credit to be given. Unsupported answers will be given zero credit. Points may be deducted when improper rounding causes significant numerical errors. All required values are given on the exam (plus some extra ones, so choose only what you need). Your calculator and one 3x5 inch card with anything you wish written on it may be brought to the exam. Additional notes or helps are prohibited.

1. (2 points) Give the molar concentrations for the following Reagent Grade chemicals as specified in the table given to you in lecture. Your number must be correct to at least two significant figures.

Sulfuric Acid	18.0 M
Ammonia (Ammonium Hydroxide)	14.5 M

2. (12 points) 500 mL of an approximately 0.03 M Nitric Acid wash solution is to be prepared using 6 M nitric acid prepared by diluting the 15 M acid found in the hood. How many milliliters of the 6 M nitric acid reagent should be added to the 500 mL of water?

Remember, start with the FINAL solution first. In this case the Final solution is 500 mL of 0.02 M HNO₃. Calculate how many moles of HNO₃ will be required, then calculate the mL of the SOURCE solution (6 M HNO₃) to use. Don't be mislead by other reagents that may be mentioned in the problem.

0.500 L	0.03 mole HNO3	L	1000 ml
	L	6 mole HNO3	L

= 2.5 mL of 6M HNO₃ required.

Remember the trick of expressing "wierd" concentrations units in a way you can do proper dimensional analysis; e.g. ppm works better as mg/Liter, and 27% Fe works better as 0.27g Fe/g unk, and GFW of Fe is 55.847 g Fe/mole Fe. We also should be used to having amounts expressed in millimoles by now, then converting to grams if we actually have to go into the balance room and weigh this AMOUNT out. Starting with the millimoles, then

2.1 mmoles Fe	1 mole Fe	55.847 g Fe	g unknown
	1000 mmole Fe	mole Fe	0.32 g Fe

= 0.366 g of unknown sample.

A classic error made on problems like this is to somehow assume that "an unknown containing chloride" is somehow a pure substance that has a formula weight and everything. Only part of it is chloride. Look to what the problem is asking you to solve:

%Cl in the sample = (grams Cl/grams sample) x 100%

We HAVE grams of sample; that is 0.3991 grams. Calculating grams of Cl will take a little more work. The one pure substance we have isolated here is the AgCl, so we must convert this to the grams of Cl that were in the sample.

23.9622 g (SGC + AgCl) - 23.4101 g (SGC alone) = 0.5521 g AgCl recovered.

The Chemical Reaction is Ag⁺ + Cl^- = AgCl (s).

0.5521 g AgCl	mole AgCl	1 mole Cl	35.453 g Cl
	143.321 g AgCl	1 mole AgCl	mole Cl

= 0.13657 g Cl (Remember to use one additonal significant figure for follow-on calculations. I still see some students rounding to 1 or 2 sig figs in their calculations. Points are taken off for this, not because of being picky, but because the final answer will be WRONG.)

Then (0.13657 g Cl/0.3991 g sample) x 100% = 34.22 %Cl (now round to 4 sig figs).

Here it is -- the deadly Relative Error Problem. As explained in lecture on several occasions, Relative errors are simple errors expressed as a percentage of the value being measured. In this case we are trying to measure a volume of about 15 mL. But since our volume change (uncertainty) is a whole 0.05 mL drop, then

(0.05 mL/20 mL) x 100% = 0.25 % error. Since 0.25 % error is greater than the 0.2% error allowed for the experiment, this is a BAD THING.

The mistake I see consistently on problems of this type is slavish adherence to a formula that is probably in all of the quant books. It is not wrong, it just doesn't cover all the cases of relative error we have discussed. The formula is:

This is a fine equation, but it doesn't apply in this problem because we already KNOW the uncertainty in our measured value, i.e. 0.05 mL. So don't look around for true and experimental values here to plug into this equation, because there are none. (Well -- 20 mL would be the True Value in this case, but there is no "Experimental Value" as such unless you make one up -- 14.95 and 15.05 would be the extremes of expected experimental values.)

A. (5 points) Mean

(By calculator) 0.12048 M (An extra significan figure is kept for follow-on calculations.)

(By calculator) 0.0110 M (A few people still did this by hand. There is no time for that!)

A. (5 points) Write the balanced dissociation reaction of Co(OH)₂ in water.

Co(OH)₂ = Co²⁺ + 2 OH^-

K_sp = [Co²⁺][OH^-]²

This is a simple solubility problem. There are no outside sources of Co²⁺ or of OH^-.

For every s moles of Co(OH)₂ that dissolve, s moles of Co²⁺ and 2s moles of OH^- appear in solution. There are no other sources of these ions. Plugging into the Ksp expression, we get

Finally, [Co²⁺] = s = 6.9x10^-6

This is a Common Ion problem, since 1.45 millimoles of Ag⁺ are left over from outside sources. Since you have already been given the final conditions here, you don't need to spend the time setting up a full Initial/Final spreadsheet.

C_Ag+ = [Ag⁺] (approximately) = 0.245 millimoles/125 mL = 0.00196 M

Then [Cl-] = Ksp/[Ag⁺] = 1.82x10^-10/0.00196 = 9.29x10^-8 M