Cracking the AP Chemistry Exam, 2019 Edition by Princeton Review

Author:Princeton Review
Language: eng
Format: epub
Publisher: Random House Children's Books
Published: 2018-11-05T16:00:00+00:00

V = 0.0374 L = 37.4 mL

(e)If both reactants are mixed in equal amounts, either value can be used to calculate the amount of precipitate that will be created.

(f) (i) Boiling the water would evaporate all the water, but it would also cause the spectator ions to no longer be dissolved in water. In addition to any PbCl2 that was created, the excess beryllium and nitrate ions would adhere to the solid, giving a falsely high molar mass and am impure product.

(ii)While decanting the solution, smaller particles of precipitate may get poured out with the water. Decanting the solution into a filtered funnel setup would be a much better option, as the filter paper would then catch all of the precipitate while allowing the water (which would carry the spectator ions) through.

2. (a)In a full redox reaction, the electrons from each half-reaction must cancel. To do that, the reduction reaction must be multiplied by two, and the oxidation reaction multiplied by five. When combining the reactions after doing so, this yields:

6H+ + 2MnO4– + 5H2O2(aq) → 2Mn2+ + 8H2O(l) + 5O2(g)

Note that 10 of the hydrogen ions cancel out when the reactions combine.

(b)A gas will be evolved as the hydrogen peroxide oxidizes into oxygen gas. Additionally, there will be a color change as the purple permanganate solution reduces into colorless manganese ions.

(c) (i) 14.03 mL – 1.52 mL = 12.51 mL

(ii) First, calculate the moles of permanganate that were titrated.

0.150 M =

n = 0.00188 mol MnO4–

Then use the stoichiometric ratios from the balanced equation to convert to moles of hydrogen peroxide.

0.00188 mol MnO4- × = 0.00470 mol H2O2

Finally, convert that to the concentration of H2O2. We will use a volume of 5.00 mL and not 50.0 mL in our calculations because we are interested in the concentration of the solution before it was diluted with water.

= 0.94 M H2O2

(d)Beakers are precise to the ones places; you cannot accurately measure out a volume in a beaker to any number of decimal places. Graduated cylinders are accurate to one decimal place, so using a beaker would reduce the number of significant figures in your answer by one.

(e) (i) If the tip of the buret is not filled prior to dispensing the permanganate solution from it, the first milliliter or so that would be dispensed is air instead of solution. Thus, the experimental volume of permanganate would be artificially high (instead of 12.51 mL of permanganate, it would have really been about 11.5 mL of permanganate with the rest being air). If the recorded volume of the permanganate is too high, that will eventually lead to a calculated [H2O2] that is also too high.

(ii) During a titration, it is best practice to regularly rinse down the sides of the flask you are titrating into, just in case any of the titrated solution splashes onto it and does not mix with the solution you are titrating into. If you do not do this, some of the

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