How to calculate the concentration of an acid in a titration

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Titration is an analytical chemistry technique used to find an unknown concentration of an analyte (the titrand) by reacting it with a known volume and concentration of a standard solution (called the titrant). Titrations are typically used for acid-base reactions and redox reactions.

Here's an example problem determining the concentration of an analyte in an acid-base reaction:

A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?

Step 1: Determine [OH-]

Every mole of NaOH will have one mole of OH-. Therefore [OH-] = 0.5 M.

Step 2: Determine the number of moles of OH-

Molarity = number of moles/volume

Number of moles = Molarity x Volume

Number of moles OH- = (0.5 M)(0.025 L)
Number of moles OH- = 0.0125 mol

Step 3: Determine the number of moles of H+

When the base neutralizes the acid, the number of moles of H+ = the number of moles of OH-. Therefore, the number of moles of H+ = 0.0125 moles.

Step 4: Determine the concentration of HCl

Every mole of HCl will produce one mole of H+; therefore, the number of moles of HCl = number of moles of H+.

Molarity = number of moles/volume

Molarity of HCl = (0.0125 mol)/(0.05 L)
Molarity of HCl = 0.25 M

Answer

The concentration of the HCl is 0.25 M.

The above steps can be reduced to one equation:

MacidVacid = MbaseVbase

where

Macid = concentration of the acid
Vacid = volume of the acid
Mbase = concentration of the base
Vbase = volume of the base

This equation works for acid/base reactions where the mole ratio between acid and base is 1:1. If the ratio were different, as in Ca(OH)2 and HCl, the ratio would be 1 mole acid to 2 moles base. The equation would now be:

MacidVacid = 2MbaseVbase

For the example problem, the ratio is 1:1:

MacidVacid = MbaseVbase

Macid(50 ml)= (0.5 M)(25 ml)
Macid = 12.5 MmL/50 ml
Macid = 0.25 M

Different methods are used to determine the equivalence point of a titration. No matter which method is used, some error is introduced, so the concentration value is close to the true value, but not exact. For example, if a colored pH indicator is used, it might be difficult to detect the color change. Usually, the error here is to go past the equivalence point, giving a concentration value that is too high.

Another potential source of error when an acid-base indicator is used is if water used to prepare the solutions contains ions that would change the pH of the solution. For example, if hard tap water is used, the starting solution would be more alkaline than if distilled deionized water had been the solvent.

If a graph or titration curve is used to find the endpoint, the equivalence point is a curve rather than a sharp point. The endpoint is a sort of "best guess" based on the experimental data.

The error can be minimized by using a calibrated pH meter to find the endpoint of an acid-base titration rather than a color change or extrapolation from a graph.

The volumes of acids and alkali solutions that react with each other can be measured by titration using a suitable indicator.

An acid-base titration is used to determine the concentration of an acid or of a base.

An acid-base titration involves the progressive addition of one reactant from a burette (buret), often the acid, to a known volume of the other reactant in a conical (erlenmeyer) flask, often the base.

An Arrhenius acid is a species that dissociates in water to produce hydrogen ions or protons, H+(aq).
An Arrhenius base is a species that dissociates in water to produce hydroxide ions, OH-(aq)

An Arrhenius acid reacts with an Arrhenius base to produce a salt and water in a neutralisation (neutralization) reaction 1:

General word equation :	Arrhenius acid	+	Arrhenius base	→	salt	+	water
General chemical equation :	H-X	+	M-OH	→	M-X	+	H-OH
HX	+	MOH	→	MX	+	H2O
Net ionic equation :	H+(aq)	+	OH-(aq)	→			H2O(l)

The equivalence point of the neutralisation titration is the point at which the moles of H+ is equal to the moles of OH-.
An indicator is used to indicate the equivalence point during a titration by changing colour2.

The titration experiment is usually conducted several times carefully and the volume of solution used from the burette (buret) recorded (known as a titre).
The volume of solution used in calculations is then the average of all these titres.

In order to calculate the concentration of an acid, c(acid), we need to know accurately:
(i) the concentration of the base used, c(base), usually in mol L-1
(ii) the volume of the base used, V(base), usually in mL which we convert to L
(iii) the volume of the acid used to neutralise the base, V(acid), usually in mL which we convert to L

In order to calculate the concentration of a base, c(base), we need to know accurately:
(i) the concentration of the acid used, c(acid), usually in mol L-1 (ii) the volume of the acid used, V(acid), usually in mL which we convert to L
(iii) the volume of the base used to neutralise the acid, V(base), usually in mL which we convert to L

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 \(\ce{NaOH}\) or \(\ce{KOH}\)). After hydrolysis is complete, the leftover base is titrated to determine how much was needed to hydrolyze the fat sample.

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

\[\text{moles acid} = \text{moles base}\nonumber \]

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.

\[\text{moles solute} = \text{M} \times \text{L}\nonumber \]

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

\[\text{M}_A \times \text{V}_A = \text{M}_B \times \text{V}_B\nonumber \]

\(\text{M}_A\) 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.

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

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 example below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.

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.

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

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 \(\ce{H_2SO_4} = 26.60 \: \text{mL}\)

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.

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

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.