KEY2CHEM

Titration

Titrations may be used to determine the concentration of an analyte in a solution can be determined by a titration. The titrant is a solution of known concentration that reacts specifically with the analyte to determine its concentration. The equivalence point of the titration occurs when the reacts completely with the titrant ($$\text{moles analyte initially present} = \text{moles titrant added}$$). The equivalence point is often detected by a color change (or change in another property), called the end point. The end point should coincide as closely as possible with the equivalence point. This observable event is called the end point of the titration.

Example 1.

In the titration of $$50.0\text{ mL}$$ of $$0.15\text{ M}$$ $$\require{mhchem}\ce{HCl}$$ with $$0.20\text{ M}$$ $$\require{mhchem}\ce{Ba(OH)2}$$, what volume of $$\require{mhchem}\ce{Ba(OH)2}$$ will have been added at the equivalence point?

$$\require{mhchem}\ce{Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + 2H2O(l)}$$

A. $$38\text{ mL}$$

B. $$19\text{ mL}$$

C. $$9.5\text{ mL}$$

Solution

B. $$19\text{ mL}$$

The volume of titrant required to reach the equivalence point can be determined from the initial quantity of analyte present, since at the equivalence point the moles of the two species are equivalent.

$$50.0\text{ mL }\require{mhchem}\ce{HCl} \times \frac{1\text{ L}}{1000\text{ mL}} \times \frac{0.15\text{ mol }\ce{HCl}}{1\text{ L }}\times \frac{1\text{ mol }\ce{Ba(OH)2}}{2\text{ mol }\ce{HCl}}\times \frac{1\text{ L }}{0.20\text{ mol }\ce{Ba(OH)2} }\times \frac{1000\text{ mL}}{1\text{ L}}= 19\text{ mL}$$

Example 2.

What is the concentration of a $$25.0\text{ mL}$$ sample of $$\require{mhchem}\ce{H2SO4}$$ analyte that requires $$44.2\text{ mL}$$  of $$0.18\text{ M}$$ $$\require{mhchem}\ce{NaOH}$$ to reach the equivalence point?

$$\require{mhchem}\ce{H2SO4(aq) + 2NaOH(aq) -> 2H2O(l) + Na2SO4(aq)}$$

A. $$0.16 \text{ M}$$

B. $$0.0040 \text{ M}$$

C. $$0.25 \text{ M}$$

Solution

A. $$0.16 \text{ M}$$

The moles of analyte can be solved using the moles of added titrant. Dividing by initial volume (in units of liters, not milliliters) yields molar concentration.

$$44.2\text{ mL }\require{mhchem}\ce{NaOH} \times \frac{1\text{ L}}{1000\text{ mL}} \times \frac{0.18\text{ mol }\ce{NaOH}}{1\text{ L }}\times \frac{1\text{ mol }\ce{H2SO4}}{2\text{ mol }\ce{NaOH}}= 4.0\times 10^{-3}\text{ mol }\ce{H2SO4}$$

$$\text{Molarity} = \frac{\text{mol }\require{mhchem}\ce{H2SO4}}{\text{ L}} = \frac{4.0\times 10^{-3}\text{ mol}}{0.0250\text{ L}} = 0.16\text{ M}$$

Example 3.

The ________________ in a titration occurs when the species of interest (the analyte) is totally consumed by the reacting species in the titrant.

A. end point

B. equivalence point

C. equilibrium point

Solution

B. equivalence point

The equivalence of the titration occurs when the analyte is totally consumed by the reacting species in the titrant. The equivalence point is often indicated by a change in a property (such as color) that occurs when the equivalence point is reached. This observable event is called the end point of the titration.