Titrations

Submitted by ChemPRIME Staff on Wed, 12/08/2010 - 23:51


A titrationThe gradual addition of one solution to another until the chemical amount of one reactant being added matches stoichiometrically the amount of another reactant in the solution initially present. is a volumetric technique in which a solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. of one reactantA substance consumed by a chemical reaction. (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence pointThe point in a titration at which the amount of one reactant being added stoichiometrically matches the amount of another reactant initially present. The end point should match the equivalence point as closely as possible. is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an indicatorA substance for which a physical property (such as color) changes abruptly when the equivalence point is reached in a titration. may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curveA graph showing the progress of a titration; for example, a plot of pH or electrical conductivity versus volume of solution added.. The amount of added titrant is determined from its concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated. and volume:

n(mol) = C(mol/L) x V(L)

and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte.

The titration process can be observed in a Chemistry Comes Alive Titration Videos, or if these videos are not available from your workstation, on YouTube Videos. Using "titration" as the keyword in YouTube finds many videos, including of an acidIn Arrhenius theory, a substance that produces hydrogen ions (hydronium ions) in aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) donor. In Lewis theory, a species that accepts a pair of electrons to form a covalent bond. titrated with baseIn Arrhenius theory, a substance that increases the concentration of hydroxide ions in an aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) acceptor. In Lewis theory, a species that donates a pair of electrons to form a covalent bond. and phenolphtalein indicator.

A measured volume of the solution to be titrated, in this case, colorless aqueousDescribing a solution in which the solvent is water. acetic acid, CH3COOH(aq) is placed in a beaker. The colorless sodium hydroxide NaOH(aq), which is the titrant, is added carefully by means of a buret. The volume of titrant added can then be determined by reading the level of liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, so precise additions of titrant can be made rapidly.

The initial reading of the buret. Placing a white card behind the buret can help the precision of the reading.

As the first few milliliters of titrant flow into the flask, some indicator briefly changes to pink, but returns to colorless rapidly. This is due to a large excess of acetic acid. The limiting reagentThe reactant (of two or more reactants) present in an amount such that it would be completely consumed if the reaction proceeded to completion. Also called limiting reactant. NaOH is entirely consumed.

The added indicator changes to pink when the titration is complete, indicating that all of the aqueous acetic acid has been consumed by NaOH(aq). The reaction which occurs is

CH3COOH(aq) + NaOH(aq) → Na+(aq) + CH3COO-(aq) + H2O(l) Eq (1)

Eventually, all the acetic acid is consumed. Addition of even a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The color change that occurs at the endpoint of the indicator signals that all the acetic acid has been consumed, so we have reached the equivalence point of the titration. If slightly more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The endpoint appears suddenly, and care must be taken not to overshoot the endpoint.

After the titration has reached the endpoint, a final volume is read from the buret. Using the initial and final reading, the volume added can be determined quite precisely:

The final reading of the buret. If the liquid is colorless, placing a white card behind the buret can aid in precise readings.

The object of a titration is always to add just the amount of titrant needed to consume exactly the amount of substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. being titrated. In the NaOH—CH3COOH reaction [Eq. (1)], the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. of CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio


\frac{n_{\text{NaOH}}\text{(added from graduated cylinder)}}{n_{\text{CH}_{\text{3}}{\text{COOH}}}\text{(initially in flask)}}=\text{S}\left( \frac{\text{NaOH}}{\text{CH}_{\text{3}}\text{COOH}} \right)


=\frac{\text{1 mol NaOH}}{\text{1 mol CH}_{\text{3}}\text{COOH}}\text{                                               (2)}



EXAMPLE 1 What volume of 0.05386 M KMnO4 would be needed to reach the endpoint when titrating 25.00 ml of 0.1272 M H2O2, given S(KMnO4/H2O2) = 2/5


Solution At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of KMnO4 which must be added:


n_{\text{KMnO}_{\text{4}}}\text{(added)}=n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\text{(in flask)}\times \text{S}\left( \frac{\text{KMnO}_{\text{4}}}{\text{H}_{\text{2}}\text{O}_{\text{2}}} \right)


The amount of H2O2 is obtained from the volume and concentration:


n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\text{(in flask)}=25.00\text{ cm}^{\text{3}}\times \text{0}\text{.1272 }\frac{\text{mmol}}{\text{cm}^{\text{3}}}=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}


Then


n_{\text{KMnO}_{\text{4}}}\text{(added)}=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}\times \frac{\text{2 mol KMnO}_{\text{4}}}{\text{5 mol H}_{\text{2}}\text{O}_{\text{2}}}\times \frac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}}


=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}\times \frac{\text{2 mmol KMnO}_{\text{4}}}{\text{5 mmol H}_{\text{2}}\text{O}_{\text{2}}}


= 1.272mmol KMnO4


To obtain VKMnO4(aq) we use the concentration as a conversion factorA relationship between two units of measure that is derived from the proportionality of one quantity to another; for example, the mass of a substances is proportional to its volume and the conversion factor from volume to mass is density.:


V_{\text{KMnO}_{\text{4}}\text{(}aq\text{)}}=\text{1}\text{.272 mmol KMnO}_{\text{4}}\times \frac{\text{1 cm}^{\text{3}}}{\text{5}\text{.386}\times \text{10}^{\text{-2}}\text{ mmol KMnO}_{\text{4}}}


= 23.62 cm3


Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4.



Titration is often used to determine the concentration of a solution. In many cases it is not a simple matterAnything that occupies space and has mass; contrasted with energy. to obtain a pure substance, weigh it accurately, and dissolve it in a volumetric flask as was done in Example 1 of Solution Concentrations. NaOH, for example, combines rapidly with H2O and CO2 from the air, and so even a freshly prepared sample of solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. NaOH will not be pure. Its weightA measure of the gravitational force on an object; directly proportional to mass. would change continuously as CO2(g) and H2O(g) were absorbed. Hydrogen chloride (HCl) is a gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container. at ordinary temperatures and pressures, making it very difficult to handle or weigh. Aqueous solutions of both of these substances must be standardized; that is, their concentrations must be determined by titration.



EXAMPLE 2 A sample of pure potassium hydrogen phthalate (KHC8H4O4) weighing 0.3421 g is dissolved in distilled water. Titration of the sample requires 27.03 ml NaOH(aq). The titration reaction is

NaOH(aq) + KHC8H4O4(aq) → NaKC8H4O4(aq) + H2O

What is the concentration of NaOH(aq) ?


Solution To calculate concentration, we need to know the amount of NaOH and the volume of solution in which it is dissolved. The former quantity could be obtained via a stoichiometric ratio from the amount of KHC8H4O4, and that amount can be obtained from the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object.


m_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{M_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}}\text{ }n_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{S\text{(NaOH/KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}\text{)}}\text{ }n_{\text{NaOH}}


n_{\text{NaOH}}=\text{3}\text{.180 g}\times \frac{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}{\text{204}\text{.22 g}}\times \frac{\text{1 mol NaOH}}{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}


=\text{1}\text{.674 }\times 10^{\text{-3}}\text{ mol NaOH}=\text{1}\text{.675 mmol NaOH}


The concentration is


c_{\text{NaOH}}=\frac{n_{\text{NaOH}}}{V}=\frac{\text{1}\text{.675 mmol NaOH}}{\text{27}\text{.03 cm}^{\text{3}}}=\text{0}\text{.06197 mmol cm}^{\text{-3}}


or      0.06197 M.



By far the most common use of titrations is in determining unknowns, that is, in determining the concentration or amount of substance in a sample about which we initially knew nothing. The next example involves an unknown that many persons encounter every day.



EXAMPLE 3 Vitamin C tablets contain ascorbic acid (C6H8O6) and a starch “filler” which holds them together. To determine how much vitamin C is present, a tablet can be dissolved in water andwith sodium hydroxide solution, NaOH(aq). The equation is


C6H8O6(aq) + NaOH(aq) → Na C6H7O6(aq) + H2O(l)

If titration of a dissolved vitamin C tablet requires 16.85 cm³ of 0.1038 M NaOH, how accurate is the claim on the label of the bottle that each tablet contains 300 mg of vitamin C?


Solution The known volume and concentration allow us to calculate the amount of NaOH(aq) which reacted with all the vitamin C. Using the stoichiometric ratio


\text{S}\left( \frac{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{NaOH}} \right)=\frac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}}


we can obtain the amount of C6H8O6. The molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. converts that amount to a mass which can be compared with the label. Schematically


V_{\text{NaOH}}\xrightarrow{c_{\text{NaOH}}}n_{\text{NaOH}}\xrightarrow{\text{S(C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}\text{/NaOH)}}n_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\xrightarrow{M_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}}\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}


\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\text{16}\text{.85 cm}^{\text{3}}\times \frac{\text{0}\text{.1038 mmol NaOH}}{\text{1 cm}^{\text{3}}}\times \frac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}}\times \frac{\text{176}\text{.1 mg }}{\text{mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}


= 308.0 mg


Note that the molar mass of C6H8O6


\frac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\frac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\times \frac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}}


=\frac{\text{176}\text{.1 g}\times \text{10}^{\text{-3}}\text{ }}{\text{10}^{\text{-3}}\text{ mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\frac{\text{176}\text{.1 mg }}{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}


can be expressed in milligrams per millimole as well as in grams per mole.



The 308.0 mg obtained in this example is in reasonably close agreement with the manufacturer’s claim of 300 mg. The tablets are stamped out by machines, not weighed individually, and so some variation is expected.