Page 235 - Theory and Problems of BEGINNING CHEMISTRY

224 SOLUTIONS [CHAP. 15

EXAMPLE 15.21. Calculate the equivalent mass of H 2 C 2 O 4 in its complete neutralization by NaOH.
−
Ans. H 2 C 2 O 4 + 2OH −→ C 2 O 4 2− + 2H 2 O
90.0g 1 mol 45.0g
=
1 mol 2 equiv 1 equiv
EXAMPLE 15.22. Calculate the molar mass of an acid with three replaceable hydrogen ions and an equivalent mass of
31.2 g/equiv, assuming complete neutralization.
Ans. 31.2g 3 equiv = 93.6g
1 equiv 1 mol 1 mol

Solved Problems

QUALITATIVE CONCENTRATION TERMS
15.1. Describe each of the solutions indicated as saturated, unsaturated, or supersaturated. (a) More solute is
added to a solution of that solute, and the additional solute all dissolves. Describe the original solution. (b)
More solute is added to a solution of that solute, and the additional solute does not all dissolve. Describe
the final solution. (c) A solution is left standing, and some of the solvent evaporates. After a time, some
solute crystallizes out. Describe the final solution. (d) A hot saturated solution is cooled slowly, and no
solid crystallizes out. The solute is a solid that is more soluble when hot than when cold. Describe the
cold solution. (e) A hot solution is cooled slowly, and after a time some solid crystallizes out. Describe
the cold solution.
Ans. (a) The original solution was unsaturated, since it was possible to dissolve more solute at that temperature.
(b) The final solution is saturated; it is holding all the solute that it can hold stably at that temperature, so not all
the excess solute dissolves. (c) The final solution is saturated; it has solid solute in contact with the solution,
and that solute does not dissolve. The solution must be holding as much as it can at this temperature. (d) The
solution is supersaturated; it is holding more solute than is stable at the cold temperature. (e) The solution
is saturated.

15.2. Ce 2 (SO 4 ) 3 is unusual because it is a solid that dissolves in water better at low temperatures than at high
temperatures. State how you might attempt to make a supersaturated solution of this compound in water
at 50 C.
◦
Ans. Dissolve as much as possible in water at 0 C. Then warm the solution carefully to 50 C. (Unless it has been
◦
◦
done before, there is no guarantee that any particular compound will produce a supersaturated solution, but
this is the way to try.)
15.3. The solubility of ethyl alcohol in water is said to be infinite. What does that mean?
Ans. It means that the alcohol is completely soluble in water no matter how much alcohol or how little water is
present.

MOLALITY
15.4. Calculate the molality of each of the following solutions: (a)2.00gC 2 H 5 OH in 30.0 g H 2 O and
(b) 3.00 g NaCl in 20.0 g H 2 O.
1 mol C 2 H 5 OH 0.0435 mol C 2 H 5 OH

Ans. (a) 2.00 g C 2 H 5 OH = 0.0435 mol C 2 H 5 OH = 1.45 m
46.0gC H 5 OH 0.0300 kg H O
2 2

1 mol NaCl 0.0513 mol NaCl
(b) 3.00 g NaCl = 0.0513 mol NaCl = 2.57 m NaCl
58.5 g NaCl 0.0200 kg H O
2
15.5. How many moles of solute are there in a 1.50 m solution containing 2.22 kg of solvent?

Ans. 2.22 kg solvent 1.50 mol solute = 3.33 mol solute
1 kg solvent