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CHEMISTRY THE CENTRAL SCIENCE

CHAPTER 4: STOICHIOMETRY


#17
The solution could contain Pb2+, the precipitate could be PbSO4. Ba2+ is out of the question because it would also precipitate as Ba(OH)2. All alkali metal compounds are soluble so that takes care of K+.

#22a
The basic species come from the induced ionization of water:
NH3(aq) + H2O --> NH4+(aq) + OH-(aq)
(NH4+ + OH- is also referred to as NH4OH although this might be just a myth)

#22b
Reactivity and acidic strength do not have to be related. The strength of an acid depends only on the level of ionization. HF has a Ka of 7.2 X 10-4, that is thousands of times weaker than what we consider strong; for example the hydronium ion (Ka = 55.5).

#22c
The ionization of sulfuric acid is actually a combination of two reactions:
H2SO4 --> H+ + HSO4-
HSO4- --> H+ + SO42-
The second ionization has a Ka of 1.2 X 10-2, whereas the first ionization is extremely large, so the first ionization is what determines which solute particles dominate.

#29a
Complete equation:
2H+(aq) + 2Br-(aq) + Ca2+(aq) + 2OH-(aq) --> Ca2+(aq) + 2Br-(aq) + 2H2O (liquid)
Net ionic:
2H+(aq) + 2OH-(aq) --> 2H2O (liquid)
or you can also write: H+(aq) + OH-(aq) --> H2O (liquid)

#29b
Complete equation:
2H+(aq) + 2ClO4-(aq) + Cu(OH)2(s) --> Cu2+(aq) + 2ClO4-(aq) + 2H2O (liquid)
Net ionic: <
2H+(aq) + Cu(OH)2(s) --> 2H2O (liquid) + Cu2+(aq)

#29c
Complete equation:
3H+(aq) + 3NO3-(aq) + Al(OH)3(s) --> Al3+(aq) + 3NO3-(aq) + 3H2O (liquid)
Net ionic:
3H+(aq) + Al(OH)3(s) --> 3H2O (liquid) + Al3+(aq)

43a
Manganese with sulfuric acid:
Mn0(s) + 2H+(aq) + SO42-(aq) --> Mn2+(aq) + H20(gas) + SO42-(aq)
Net ionic:
Mn0(s) + 2H+(aq) --> Mn2+(aq) + H2(gas)

43b
Chromium with hydrobromic acid at 600-700 0C:
2Cr0(s) + 6H+(aq) + 6Br-(aq) --> 2Cr3+(aq) + 3H2(gas) + 6Br-(aq)
Net ionic:
Cr0(s) + 6H+(aq) --> Cr3+(aq) + 3H2(gas)

43c
Tin and hydrochloric acid:
Sn0(s) + 2H+(aq) + 2Cl-(aq) --> Sn2+(aq) + H2(gas) + 2Cl-(aq)
Net ionic:
Sn0(s) + 2H+(aq) --> Sn2+(aq) + H2(gas)

43d
Aluminum and formic acid:
2Al0(s) + 6HCHO2(aq) --> 2Al3+(aq) + 3H2(gas) + 6CHO2-(aq)
Net ionic:
2Al0(s) + 6HCHO2(aq) --> 2Al3+(aq) + 3H2(gas) + 6CHO2-(aq)

61a
125 mL of 0.150 M sucrose:
In 125 mL of water dissolve [0.150 mol/L] X [125 X 10-3 L] = 0.01875 mol of sucrose
This means [0.01875 mol of sucrose] X [342.3 g/mol] = 6.42 g of sucrose

61b
2.00 L of 1.5 M --> 400.0 mL of 0.100 M
We need [0.100 mol/L] X [400 X 10-3 L] = 0.0400 mol of sucrose
That can be obtained from [0.0400 mol sucrose] / [1.50 mol sucrose/L] = 0.0267 L of the original 1.5 M solution, then dilute to 400.0 mL volume.

71
The reaction:
Ca(OH)2 + 2HBr --> 2H2O + CaBr2
From stoichiometry: Mol of Ca(OH)2 = 2 HBr
Mol of HBr = [5.00 X 10-2 mol/L] X [48.80 X 10-3 L] = 0.00244 mol HBr
Mol of Ca(OH)2 = [0.00244 mol] / 2 = 0.00122 mol Ca(OH)2
Molarity of the Ca(OH)2 solution = [0.00122 mol Ca(OH)2] / [100 X 10-3 L] = 0.0122 M
Grams of Ca(OH)2 = [0.00122 mol of Ca(OH)2] X [74.0 g/mol] = 0.09028 g
The solubility is 0.09028 g of Ca(OH)2/100 mL of solution.

73a
Ni2+(aq) + SO42-(aq) + 2K+(aq) + 2OH-(aq) --> Ni(OH)2(s) + SO42-(aq) + 2K+(aq)

73b
The precipitate is Ni(OH)2

73c
Mol of KOH = [0.200 mol/L] X [100 X 10-3 L] = 0.0200 mol of KOH
Mol of NiSO4 = [0.150 mol/L] X [200 X 10-3 L] = 0.0300 mol of NiSO4
The product requires two OH- for every Ni2+. Compare one half of 0.0200 with 0.0300 and KOH is limiting.
Moles of precipitate = [0.0200 mol / 2] = 0.0100 mol
Grams of precipitate = [0.0100 mol] X [92.7 g/mol] = 0.927 grams of Ni(OH)2
Leftover = 0.0300 mol 0.0100 mol = 0.0200 mol of NiSO4
Total volume of solution = [200 X 10-3 L] + [100 X 10-3 L] = 300 X 10-3 L
Concentration of Ni2+ = 0.0200 mol / 300 X 10-3 L = 0.0667 M
Concentration of K+ = [0.0200 mol / 300 X 10-3 L = 0.0667 M
Concentration of SO42-= [0.0300 mol] / 300 X 10-3 L = 0.100 M





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