_2_ HBr(aq) + ___ Ca(OH)2 (aq) → CaBr2 (aq) + 2H2O(aq)



PRACTICE EXAM B Answers – 10/3/06

1. What are the strong acids? Strong bases? How do you know what are weak acids? What are the weak bases that we should know?

The strong acids were given in lecture: HCl, HBr, HI, HNO3, H2SO4, and HClO4. Strong bases are anything soluble with the hydroxide ion (OH-) in it. Weak acids will be anything not on your list of strong acids (ex. HF). The only weak base mentioned in class was NH3.

2. a. Is this reaction a neutralization reaction? Also, what are the ionic and net ionic equations? Yes, it is a neutralization—you ended up with a salt (ionic compound) and water, so it is more “neutral” now.

_2_ HBr(aq) + ___ Ca(OH)2 (aq) → CaBr2 (aq) + 2H2O(l)

Ionic: 2H+(aq) + 2Br-(aq) + Ca2+(aq) + 2OH-(aq) → Ca2+(aq) + 2Br-(aq) + 2H2O(l)

Net Ionic: H+(aq) + OH-(aq) → H2O(l)

Note that the "2" coefficients cancelled out! That is because the ratio for the net ionic was 2:2:2, which can be simplified to 1:1:1.

b. If given 2.0 L of 0.45 M HBr and 1.8 L of 0.16 M Ca(OH)2, which is the limiting reactant? Well, M = molarity which is moles/L of solution. So:

2.0 L HBr x 0.45 mol HBr = 0.90 mol HBr

1 L

1.8 L Ca(OH)2 x 0.16 mol Ca(OH)2 = 0.29 mol Ca(OH)2

1 L

Since we need 2 mol HBr for every 1 mol Ca(OH)2, we can see that there is more than 2 mol of HBr for every 1 of Ca(OH)2 (0.29 mol x 2 = needed HBr = 0.58 mol HBr needed). HBr is in excess, and Ca(OH)2 is therefore the limiting reactant.

c. What is the conjugate base to the acid in the above reaction? Is it a strong or weak conjugate base? What is the conjugate acid to the base above? Is it strong or weak?

Conjugate bases are the remaining anion when H+ is given off an acid. Conjugate acids are the remaining cations when OH- is given off of a base. Therefore, the conjugate base is Br-, and it is a weak base because HBr was a strong acid (opposite relationship). The conjugate acid is Ca2+, and it is a weak acid because Ca(OH)2 is a strong base.

3. a. If the concentration of an acid’s protons is 8.5x10-3 M, what is the pH? Is it a strong or weak acid? pH = -log [H+] = -log (8.5x10-3 M) = 2.1

This acid is a strong acid because it has a very low pH (on a scale of 0-14, with 14 being extremely basic and 0 being extremely acidic).

b. If the concentration of a base’s hydroxide ions is 5.0x10-6 M, what is the pH? Is it a strong or weak base? pOH = -log [OH-] = -log (5.0x10-6 M) = 5.3

pH + pOH = 14, therefore: 14-pOH = pH = 14-5.3 = 8.7

This is a weak base because the pH value is basic (above 7) but is not that high.

4. Assign oxidation numbers for each of the elements in the following compounds.

a. NO2

N: +4 O: -2

b. Na2SO4

Na: +1 S: +6 O: -2

When you see an ionic compound, split it up into its ions. Therefore, the entirety of SO4 has a –2 charge. Start with O (which has a number of –2) and then calculate the number for sulfur.

c. HPO42-

H: +1 P: +5 O: -2

d. BrO-

Br: +1 O: -2

5. Which one is oxidized in the following reaction? Which one is reduced? What are the oxidizing and reducing agents?

a. Ag(s) + NO3-(aq) → Ag+(aq) + NO2(g)

The oxidation number for silver went from 0 to +1. Ag was oxidized and is the reducing agent.

The oxidation number of N goes from +5 to +4, so it gained an electron and was thus reduced and it is the oxidizing agent.

b. Mg(s) + VO43-(aq) → Mg2+(aq) + V2+(aq)

The oxidation number of Mg went from 0 to +2, so it was oxidized and is the reducing agent.

The oxidation number for V went from +5 to +2, so it was reduced and is the oxidizing agent.

c. What are the best oxidizing and reducing agents?

The best oxidizing agents are non-metals, such as Nitrogen above, and higher charged B (transition) metals, such as Vanadium above.

The best reducing agents are metals, such as Silver (Ag) and Magnesium above, and lower charged B (transition) metals.

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