http://www.ck12.org Chapter 21. Acids and Bases
We can then set the moles of acid equal to the moles of base.
MA×VA= MB×VB
MAis the molarity of the acid, while MBis the molarity of the base. VAand VBare the volumes of the acid and base,
respectively.
Suppose that a titration is performed and 20.70 mL of 0.500 M NaOH is required to reach the end point when titrated
against 15.00 mL of HCl of unknown concentration. The above equation can be used to solve for the molarity of the
acid.
MA=
MB×VB
VA
=
0 .500 M× 20 .70 mL
15 .00 mL= 0 .690 M
The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required
to reach the equivalence point.
The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. Sample
Problem 21.6 demonstrates a titration problem for which this is not the case.
Sample Problem 21.6: Titration
In a titration of sulfuric acid with sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL
of the H 2 SO 4 solution. Calculate the molarity of the sulfuric acid.
Step 1: List the known values and plan the problem.
Known
- molarity of the NaOH solution = 0.250 M
- volume of the NaOH solution = 32.20 mL
- volume of the H 2 SO 4 solution = 26.60 mL
Unknown
- molarity of the H 2 SO 4 solution =?
Equation: H 2 SO 4 (aq)+2NaOH(aq)→Na 2 SO 4 (aq)+2H 2 O(l)First, determine the moles of NaOH that were consumed in the reaction. From the mole ratio, calculate the moles of
H 2 SO 4 that reacted. Finally, divide the moles of H 2 SO 4 by the volume of the sample of acid to get the molarity.
Step 2: Solve.
mol NaOH=M×L= 0 .250 M× 0 .03220 L= 8. 05 × 10 −^3 mol NaOH
8. 05 × 10 −^3 mol NaOH×1 mol H 2 SO 4
2 mol NaOH
= 4. 03 × 10 −^3 mol H 2 SO 4
4. 03 × 10 −^3 mol H 2 SO 4
0 .02660 L= 0 .151 M H 2 SO 4
Step 3: Think about your result.
The volume of H 2 SO 4 required is smaller than the volume of NaOH because of the two hydrogen ions contributed
by each molecule.