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# Titration Calculations

## Calculations used to determine analyte concentrations in titration experiments.

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Titration Calculations

Credit: User:Phanton/Wikipedia
Source: http://commons.wikimedia.org/wiki/File:Decorative_Soaps.jpg

The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product. The fat is heated with a known amount of base (usually NaOH or KOH). After hydrolysis is complete, the left-over base is titrated to determine how much was needed to hydrolyze the fat sample.

### Titration Calculations

At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.

moles acid=moles base

Recall that the molarity  (M)\begin{align*}(M)\end{align*} of a solution is defined as the moles of the solute divided by the liters of solution (L)\begin{align*}(L)\end{align*} . So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.

moles solute=M×L

We can then set the moles of acid equal to the moles of base.

MA×VA=MB×VB

MA\begin{align*}M_A\end{align*}  is the molarity of the acid, while  MB\begin{align*}M_B\end{align*} is the molarity of the base.  VA\begin{align*}V_A\end{align*} and  VB\begin{align*}V_B\end{align*} are 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×VBVA=0.500 M×20.70 mL15.00 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. The sample problem below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.

#### Sample Problem: Titration

In a titration of sulfuric acid against sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of H 2 SO 4 . Calculate the molarity of the sulfuric acid.

Step 1: List the known values and plan the problem.

Known

• molarity NaOH = 0.250 M
• volume NaOH = 32.20 mL
• volume H 2 SO 4 = 26.60 mL

Unkonwn

• molarity H 2 SO 4 = ?

equationH2SO4(aq)+2NaOH(aq)Na2SO4(aq)+2H2O(l)

First determine the moles of NaOH in the reaction. From the mole ratio, calculate the moles of H 2 SO 4 that reacted. Finally, divide the moles H 2 SO 4 by its volume to get the molarity.

Step 2: Solve.

mol NaOH=M×L=0.250 M×0.03220 L=8.05×103 mol NaOH8.05×103 mol NaOH×1 mol H2SO42 mol NaOH=4.03×103 mol H2SO44.03×103 mol H2SO40.02660 L=0.151 M H2SO4

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.

#### Summary

• The process of calculating concentration from titration data is described and illustrated.

#### Practice

Do the problems at the link below:

#### Review

Questions

1. What assumption is made about the amounts of materials at the neutral point?
2. What is different about the calculation using sulfuric acid?
3. Why is the mole ratio important?