Volumetric (Titrimetric) Analysis
General Principles
In titrimetric analysis volumetrically measures the amount of reagent, often called a
titrant, required to complete a chemical reaction with the analyte. A generic chemical
reaction for titrimetric analysis is
where a moles of analyte A contained in the sample reacts with t moles of the titrant T in the titrant solution.
The reaction is generally carried out in a flask containing the liquid or dissolved sample. Titrant solution is volumetrically delivered to the reaction flask using a burette. Delivery of the titrant is called a titration. The titration is complete when sufficient titrant has been added to react with all the analyte. This is called the equivalence point
An indicator is often added to the reaction flask to signal when all of the analyte has reacted. The titrant volume where the signal is generated is called the end point. The equivalence and end points are rarely the same.
Successful Titrimetric Analysis
A few rules of thumb for designing a successful titration are:
Types of Chemistry
Although any type of chemical reaction may be used for titrimetric analysis, the reactions most often used fall under the categories of
Bronsted AcidBase
OxidationReduction
Precipitation
Complex Formation
Lewis AcidBase chemistry is often involved in precipitation and complex formation chemistry (especially when using or analyzing for transition metals).
Steps in a Titration
Titrimetric analysis generally involve the following steps:
Sample Calculations
Titrimetric analyses are often performed with singlepoint standardization. Standardization of a base titrant solution, used to determine an amount of acid in a sample, is often performed using a known, dry weight of potassium hydrogen phthalate (KHP) as the standard. There is one proton, or equivalent, per mole of KHP. The base reaction is
The titer value is determined from base titrant standardization. The titer value is the amount of acid neutralized per mL titrant solution.
For example, if 3 standards, each with 2.50 mmole of KHP required 25.3 mL, 24.9 mL, and 24.8 mL of titrant, respectively, then the titer value for the titrant would be the moles of acid neutralized divided by the average titrant volume. In this case the average is 25.0 mL and the standard deviation is s = 0.265 mL
The sample (liquid) is now analyzed. Triplicate analyses of 25.0 mL sample aliquots of the acid of unknown strength produced titrant volumes of 15.4 mL, 14.8 mL, and 14.8 mL, for average and standard deviation values of 15.2 mL and 0.346 mL, respectively.
The moles unknown acid protons reacted is the titer value multiplied by the average volume
The acid concentration is obtained as formal proton concentration using the sample volume. In this case
Error Analysis
The standard deviation of this result is a combination of that for the onestep standardization and sample titration steps. To see where this error comes from, the above procedure may be combined into a single equation
It is assumed that the KHP weights and sample volumes are accurately known. Errors are associated only with estimates of the two titrant volumes, given in bold. Since the combined equation uses division of one uncertain result by another, the errors are propagated as relative standard deviations
The relative errors due to moles KHP for the standard and mL sample are the zeros since these measurements are assumed to be precise and accurate.
The standard deviation, in formal concentration units, is s_{F}=( s_{F} /F)´ F=0.00153 The standard deviation may be used to report the value using the 95% Student’s t value for 31=2 degrees of freedom of 4.303
Notice that the results are rounded to the third place because of the third place error of the confidence interval.
Titration Curves
Titration curves are constructed by plotting –log_{10} concentration, or pX, of the analyte versus the volume of titrant added to the reaction solution. For example, for acidbase titrations, the pH or pOH is plotted versus base or acid titrant volume. Titration curves may be plotted from experimental data to increase the precision of the resulting unknown determination. Similarly, theoretical titration curves may be used to investigate the feasibility of a titration.
Consider the case of calculating the titration curve for 25 mL 0.1 F NaI analyte being titrated with 0.05 F AgNO_{3} titrant. This will prove to be a feasible titration if a suitable indicator can be found.
NaI and AgNO_{3} are both moderately soluble salt. The precipitation titration proceeds via the chemical reaction
K_{sp} for AgI is only 8.3´ 10^{17}. To generate the titration curve, the concentration of I^{} as a function of titrant volume needs to be found. This may be done by calculating [I^{}] at a number of volumes, and drawing a smooth curve through the resulting points on the graph.
The general approach is to calculate [I^{}] initially, at the equivalence point, half way to the equivalence point, and then with excess Ag^{+}. The [I^{}] is found using the usual steps for the common ion solubility effect; first assume as much as possible insoluble AgI salt is formed; then, using the initial/change/final table procedure, calculate the equilibrium [I^{}].
At the beginning of the titration, no titrant has been added, and the [I^{}] is just that from 0.1 F NaI solution. Specifically, [I^{}]=0.1 M. Thus, the initial pI is –log_{10}[I^{}] = 1.
The equivalence point is the titrant volume required to react with all I^{} initially in the sample. Initially, there is
Since one mole Ag^{+} is required for each mole of I^{}, 2.50 mmole Ag^{+} is needed. The volume of titrant needed to reach the equivalence point is
Assuming that all of I^{} is initially as AgI
AgI (s) Û 
Ag^{+} (aq) 
I^{} (aq) 

Initial  s 
0 
0 
Change  x 
+x 
+x 
Final  s 
x 
x 
Using the solubility product expression
In general, the pX at the equivalence point of a precipitation titration is directly related to the pK_{sp}. In this case, pX= pK_{sp}/2.
Half way to the equivalence point, half of the equivalence point volume of titrant has been added, i.e., V_{titrant}=50.0 mL/2 or 25.0 mL. At this point, only a maximum of one half of the initial I^{} or 1.25 mmole, can react to form AgI. The minimum, initial concentration of I^{} must take into account dilution due to added titrant. The total volume is 25.0 mL sample + 25.0 mL titrant = 50.0 mL. The initial [I^{}] is thus
Assuming that all of I^{} is initially as AgI, the initial/change/final table at this point is
AgI (s) Û 
Ag^{+} (aq) 
I^{} (aq) 

Initial  s 
0 
0.025 
Change  x 
+x 
+x 
Final  s 
x 
0.025+x 
Substitution into the solubility product equation yields
Assuming that x is much less than 0.025
The assumption is apparently correct. Since x is very small relative to 0.025 M, the iodide concentration very close to this value. Subsequently, pI=1.6 half way to the equivalence point.
The point were halfagain as much titrant is added past the equivalence point is used as the final point. In other words, a titrant volume of 75.0 mL is used. At this point we assume that all iodide is initially as insoluble AgI salt. The total silver ion added to the solution is
The amount of excess silver ion is the that amount added, less the amount reacted
The total solution volume is volume is 25.0 mL sample + 75.0 mL titrant = 100.0 mL. The initial Ag^{+} concentration is
The salt partially dissolves
AgI (s) Û 
Ag^{+} (aq) 
I^{} (aq) 

Initial  s 
0.0125 
0 
Change  x 
+x 
+x 
Final  s 
0.0125+x 
x 
Substitution of these results into the solubility equation
Assuming that x is much less than 0.0125
The assumption is apparently correct. Since [I^{}]=x, the pI=14.18
Final Product
The points calculated above are plotted as diamonds in the figure shown here. One might notice that the four points do not unambiguously give the smooth line plotted. With experience, the shape of this and similar titration curves will be obvious and you will be able to estimate the curve based on only a few points. So how does one make these "pretty pictures" with smooth lines? Clearly, I didn't draw them in. The smooth line was generated using a spreadsheet program. A bit about how one may we obtain the smooth line is discussed below.
Titration Equation
The general answer is that one uses an equation that describes the iodide concentration as a function of added titrant. Sometimes this equation is hard to come by. However, the derivation is straightforward in this case.
At any point along the titration, the sum of all charges in solution must sum to zero. Net charge cannot be generated by addition of one chargeneutral solution to another. The algebraic equation representing this fact is called a charge balance equation. In this case, the charge balance equation is
Another necessary equation is the solubility product
Since the goal is to calculate [I^{}] one solves the solubility product equation for [Ag^{+}] and substitutes this into the charge balance equation. One obtains a quadratic equation after rearranging
The solution is the usual quadratic equation root
[Na^{+}] and [NO_{3}^{}] concentrations are easily obtained from dilution equations since these species do not participate in the chemical reaction.
The calculations are easily performed using a spreadsheet program.
This page was created by Professor Stephen Bialkowski, Utah State University.
Last Updated Wednesday, September 28, 2005