Page 177 - Theory and Problems of BEGINNING CHEMISTRY

166 MOLARITY [CHAP. 11

toward the end, until a permanent color change takes place. At that point, the endpoint has been reached, and
the titration is complete. The HCl has just been used up. (2) The purpose of doing a titration is to determine the
concentration of a solution. If the concentration of 1 L of NaOH is to be determined, only a small portion of it is
used in the titration. The rest has the same concentration, and although the part used in the titration is no longer
useful, the concentration of most of the solution is now known.

EXAMPLE 11.12. What is the concentration of a Ba(OH) 2 solution if it takes 43.12 mL to neutralize 25.00 mL of
2.000 M HCl?
Ans. The number of moles of HCl is easily calculated.

2.000 mmol 2.000 mol

25.00 mL = 50.00 mmol or 0.025 00 L = 0.050 00 mol HCl
1mL 1L
The balanced chemical equation shows that the ratio of moles of HCl to Ba(OH) 2 is2:1.
2 HCl + Ba(OH) 2 −→ BaCl 2 + 2H 2 O

1 mmol Ba(OH) 2 1 mol Ba(OH) 2
50.00 mmol HCl or 0.050 00 mol HCl
2 mmol HCl 2 mol HCl
= 25.00 mmol Ba(OH) 2 = 0.025 00 mol Ba(OH) 2
The molarity is given by
25.00 mmol Ba(OH) 2 0.025 00 mol Ba(OH) 2
= 0.5798 M or = 0.5798 M Ba(OH) 2
43.12 mL 0.043 12 L
Note that you cannot calculate this concentration with the equation

M 1 V 1 = M 2 V 2 (limited applications)
where M is molarity and V is volume. This equation is given in some texts for simple dilution problems. The equation
reduces to moles 1 = moles 2 , which is not true in cases in which there is nota1:1 ratio in the balanced chemical
equation.

11.4. STOICHIOMETRY IN SOLUTION
With molarity and volume of solution, numbers of moles can be calculated. The number of moles may be
used in stoichiometry problems just as moles calculated in any other way are used. Also, the number of moles
calculated as in Chap. 10 can be used to calculate molarities or volumes of solution.

EXAMPLE 11.13. Calculate the number of moles of AgCl that can be prepared by mixing 0.740 L of 1.25 M AgNO 3 with
excess NaCl.

Ans. NaCl + AgNO −→ AgCl(s) + NaNO 3
3
(0.740 L)(1.25 mol AgNO /L) = 0.925 mol AgNO 3
3
(0.925 mol AgNO )(1 mol AgCl/1 mol AgNO ) = 0.925 mol AgCl
3
3
EXAMPLE 11.14. Calculate the number of moles of AgCl that can be prepared by mixing 0.740 L of 1.25 M AgNO 3 with
1.50Lof0.900 M NaCl.
Ans. NaCl + AgNO −→ AgCl(s) + NaNO 3
3
(0.740 L)(1.25 mol AgNO /L) = 0.925 mol AgNO present
3
3
(1.50 L)(0.900 mol NaCl/L) = 1.35 mol NaCl present
AgNO 3 is in limiting quantity, and the problem is completed just as the prior example was done.