Acids and Bases: Titration Example Problem

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:

Titration Problem Step-by-Step Solution

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⁺

Read More

Solve Acid-Base Titration Calculations

By Anne Marie Helmenstine, Ph.D.

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.

Another Solution Method

The above steps can be reduced to one equation:

M_acidV_acid = M_baseV_base

where

M_acid = concentration of the acid
V_acid = volume of the acid
M_base = concentration of the base
V_base = 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:

M_acidV_acid = 2M_baseV_base

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

M_acidV_acid = M_baseV_base

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

Error in Titration Calculations

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.