Download Procedure for solving acid/base titration problems when the mole

Procedure for solving acid/base titration problems when the mole ratio of acid to base is 1:1.
1. Begin with a correctly written balanced equation for the neutralization reaction.
2. Find the number of moles of the known standard (often called the titrant) used to neutralize the unknown.
Use molarity and volume in liters. (M x V = # moles).
3. Use the mole ratio from the balanced equation to find the number of moles of the unknown neutralized.
When the mole ratio of acid to base is 1:1, the number of moles of acid equals the number of moles of
base.
4. To find the molarity of the unknown… number of moles ÷ volume of unknown in liters.
Example Titration Problem with a 2:1 Acid/Base Mole Ratio.
A 21.62 mL sample of Ca(OH)2 solution was titrated with 0.2545 M HCl. 45.87 mL of the acid was required to
reach the endpoint of the titration (the neutralization point)
(a) What is the equation of the reaction?
(b) What was the molarity of calcium hydroxide solution?
Solution:
1) First, write the balanced chemical equation:
2HCl + Ca(OH)2 ---> CaCl2 + 2H2O
2) Second, the final solution to problem involves several steps:
Using molarity and volume (in liters) find moles of HCl…
(0.2545 mol/L) (0.04587 L) = 0.011674 mol HCl (molarity x volume in liters = #moles)
Use the mole ratio of HCl to Ca(OH)2 (2 : 1) to find numbers of moles of base required to neutralize the acid.
This can be done with a proportion or via dimensional analysis.
Using a proportion…
x= 0.005837 mol Ca(OH)2
Using dimensional analysis…
0.011674 mol HCl
x
1 Ca(OH)2
2 HCl
=
0.005837 mol of Ca(OH)2
Finally,
Use the moles of base and the volume of base (in liters) to calculate the molarity of the Ca(OH)2
molarity of Ca(OH)2 ---> 0.005837 mol / 0.02162 L = 0.2700 M (to four sig figs)