Page 181 - Theory and Problems of BEGINNING CHEMISTRY

170 MOLARITY [CHAP. 11

11.20. What concentration of salt is obtained by mixing 20.0 mL of 3.0 M salt solution with 30.0 mL of 2.0 M
salt solution and diluting with water to 100.0 mL?
Ans. The final concentration is the total number of moles divided by the total number of liters. The volume is
100.0 mL. Since there is no solute in the water, the total number of moles is the same as that in the prior
problem. The concentration is 120 mmol/100.0 mL = 1.2 M. The concentration is lower than that in Problem
11.19 despite the same number of moles of solute, because of the greater volume.

11.21. Calculate the concentration of sugar in a solution prepared by mixing 3.0 L of 2.0 M sugar with 2.5 L of
1.0 M salt.
Ans. The sugar concentration is given by dividing the number of moles of sugar by the total volume. The number of
moles of salt makes no difference in this problem because the problem does not ask about salt concentration
and the salt does not react. This is simply a dilution problem for the sugar.
(3.0L)(2.0 mol/L) = 6.0 mol sugar
6.0 mol sugar
= 1.1 M sugar
5.5L

TITRATION
11.22. (a) A solid acid containing one hydrogen atom per molecule is titrated with 1.000 M NaOH. If 27.21 mL
of base is used in the titration, how many moles of base is present? (b) How many moles of acid? (c)If
the mass of the acid was 3.494 g, what is the molar mass of the acid?
Ans. (a) (27.21 mL)(1.000 mmol/mL) = 27.21 mmol base
(b) Since the acid has one hydrogen atom per molecule, it will react in a 1:1 ratio with the base. The
equation might be written as
HX + NaOH −→ NaX + H 2 O
where X stands for the anion of the acid, whatever it might be (just as x is often used for an unknown
in algebra). The quantity of acid is therefore 27.21 mmol, or 0.027 21 mol.

3.494 g
(c) Molar mass = = 128.4 g/mol
0.027 21 mol

11.23. A 25.00-mL sample of 1.000 M HCl is titrated with 31.72 mL of NaOH. What is the concentration of
the base?
Ans. (25.00 mL HCl)(1.000 mmol/mL) = 25.00 mmol HCl
Since the reagents react in a 1:1 ratio, there is 25.00 mmol NaOH in the 31.72 mL of base.
25.00 mmol NaOH
= 0.7881 M NaOH
31.72 mL

11.24. A 25.00-mL sample of 2.000 M H 2 SO 4 is titrated with 16.54 mL of NaOH until both hydrogen atoms of
each molecule of the acid are just neutralized. What is the concentration of the base?
Ans. (25.00 mL H 2 SO 4 )(2.000 mmol/mL) = 50.00 mmol H 2 SO 4
H 2 SO 4 + 2 NaOH −→ Na 2 SO 4 + 2H 2 O

2 mmol NaOH
50.00 mmol H 2 SO 4 = 100.0 mmol NaOH
1 mmol H 2 SO 4
100.0 mmol NaOH
= 6.046 M NaOH
16.54 mL