# 7.19: Titration Calculations


## Titration Calculations

For a neutralization reaction with a 1:1 stoichiometric ratio between the acid and the base, at the equivalence point in a neutralization, the moles of acid are equal to the moles of base.

$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)$

$\text{moles acid} = \text{moles base}$

Recall that the molarity $$\left( \text{M} \right)$$ of a solution is defined as the moles of the solute divided by the liters of solution $$\left( \text{L} \right)$$. So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.

at the equivalence point : $\text{moles solute} = \text{M} \times \text{L}$

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

$\text{M}_A \times \text{V}_A = \text{M}_B \times \text{V}_B$

$$\text{M}_A$$ is the molarity of the acid, while $$\text{M}_B$$ is the molarity of the base. $$\text{V}_A$$ and $$\text{V}_B$$ are the volumes of the acid and base, respectively.

Suppose that a titration is performed and $$20.70 \: \text{mL}$$ of $$0.500 \: \text{M} \: \ce{NaOH}$$ is required to reach the end point when titrated against $$15.00 \: \text{mL}$$ of $$\ce{HCl}$$ of unknown concentration. The above equation can be used to solve for the molarity of the acid.

$\text{M}_A = \frac{\text{M}_B \times \text{V}_B}{\text{V}_A} = \frac{0.500 \: \text{M} \times 20.70 \: \text{mL}}{15.00 \: \text{mL}} = 0.690 \: \text{M}$

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.  When the formula of the acids or the base determines a different molar ratio, then the equation above cannot be applied. In this case, we can use traditional stoichiometric calculations. The example below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide, which react in a 1:2 moles ratio.

Example $$\PageIndex{1}$$

In a titration of sulfuric acid against sodium hydroxide, $$32.20 \: \text{mL}$$ of $$0.250 \: \text{M} \: \ce{NaOH}$$ is required to neutralize $$26.60 \: \text{mL}$$ of $$\ce{H_2SO_4}$$. Calculate the molarity of the sulfuric acid.

$\ce{H_2SO_4} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Na_2SO_4} \left( aq \right) + 2 \ce{H_2O} \left( l \right)$

Solution

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

• Molarity $$\ce{NaOH} = 0.250 \: \text{M}$$
• Volume $$\ce{NaOH} = 32.20 \: \text{mL}$$
• Volume $$\ce{H_2SO_4} = 26.60 \: \text{mL}$$

Unknown

• Molarity $$\ce{H_2SO_4} = ?$$

First determine the moles of $$\ce{NaOH}$$ in the reaction. From the mole ratio, calculate the moles of $$\ce{H_2SO_4}$$ that reacted. Finally, divide the moles of $$\ce{H_2SO_4}$$ by its volume to get the molarity.

Step 2: Solve.

\begin{align} &\text{mol} \: \ce{NaOH} = \text{M} \times \text{L} = 0.250 \: \text{M} \times 0.03220 \: \text{L} = 8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \\ &8.05 \times 10^{-3} \: \text{mol} \: \ce{NaOH} \times \frac{1 \: \text{mol} \: \ce{H_2SO_4}}{2 \: \text{mol} \: \ce{NaOH}} = 4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4} \\ &\frac{4.03 \times 10^{-3} \: \text{mol} \: \ce{H_2SO_4}}{0.02660 \: \text{L}} = 0.151 \: \text{M} \: \ce{H_2SO_4} \end{align}

The volume of $$\ce{H_2SO_4}$$ required is smaller than the volume of $$\ce{NaOH}$$ because of the two hydrogen ions contributed by each molecule.