In freshman chemistry, we treat titration this way. A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution. Typically, the titrant (the known solution) is added from a buret to a known quantity of the analyte (the unknown solution) until the reaction is complete. Knowing the volume of titrant added allows the determination of the concentration of the unknown. Often, an indicator is used to usually signal the end of the reaction, the endpoint.
In hospital and medical labs, automated titration equipments are used. The following sites have some information regarding the automated titrator:
For acid-base titration, a modern lab will usually monitor titration with a pH meter which is interfaced to a computer, so that you will be able to plot the pH or other physical quantities versus the volume that is added.
In this module we simulate this experiment graphically without using chemicals. A program that simulates titrations of strong acids and strong bases is very easy, because the calculation of pH in this experiment is very simple.
0 1.0 0.0
1.0 9/11 0.087
2.0 8/12 0.176
5.0 5/15 0.477
8.0 2/18 0.954
9.0 1/19 1.279
9.3 0.7/19.3 1.440
9.5 0.5/19.5 1.591
9.7 0.3/19.7 1.817
9.8 0.2/19.8 2.0
9.9 0.1/19.9 2.300
9.95 0.05/19.95 2.60
pH = 7
NaCl neutral salt
10.05 0.05/20.05 11.397
10.10 0.5/20.1 11.697
11.0 1/21 12.678
15.0 5/25 13.301
20.0 20/30 13.924
The amount of acid present = Va*Ca
= 10.0 mL * 1.0 mol/1000 mL
= 10 mmol (mili-mole)
The amount of base NaOH added = Vb*Cb
The amount of acid left = Va*Ca - Vb*Cb
The concentration of acid and thus [H+]
= [Va*Ca - Vb*Cb] / (Va + Vb)
With the above formulation, we can built a table for various values as shown on the right.
Working to learn
Plot the titration curve on a graph based on the data.
Answer the following questions.
At equivalent point, why is pH=7? What formula is used to calculate pH?
Why does pH change rapidly at the equivalent point?
Sketch titration curves when the concentrations of both acids and bases are 0.10, 0.0010 and 0.000010 M? What can you conclude from these sketches?
What are [Na+] and [Cl-] at the following points: initially (before any base is added), half-equivalent point; equivalent point, after 10.5 mL NaOH is added, after 20.0 mL NaOH is added? Well, when you have acquired the skill to calculate the pH at any point during titration, you may write a simulation program to plot the titration curve. Calculations for strong-acid_strong-base titration are simple, but when weak acid or base are involved, the calculations are somewhat more complicated. However, we are interested in this area and some simulation programs are available on the internet.
Yue-Ling Wong has given a Java interactive titration simulation. His website is rather fun to play with and it is nicely done. In fact, the design of Wong's Java interactive simulation is very much like the design of the DOS CACT version.
Using a computer we are able to simulate the titrations of weak acids and strong bases, or strong acids and weak bases. Calculation of pH in the titration of weak acids with weak bases is more difficult. However, let the complexity bother you no more, since you can simply have fun testing the computer model.
In a titration experiment, the amount to add from the buret depends on the condition at the time. At the start, you may add large amount before observing much pH change, but when the titration is at its end or equivalent point, you would like to ad the titrant slowly. In the simulation, you control the rate of titration.
While trying the Java simulation of titration, please answer the following questions.
Answer pH=-1, 0, 1, 2, and 3
No calculators should be used.
Answer [H+]=0.333; pH=0.477
Do not forget the dilution factor.
What about [Cl-]?
Answer [H+]=1e-7; pH=7
This is only a theoretical value.
Answer [Ca+] = [Cl-] = 0.5
Note that [H+] is balanced by [OH-], amd [Na+] is balanced by [Cl-].
Most students usually forget the salt resulting from the titration.
Answer [H+]=0.00424; pH=2.37
The approximation [H+]=(C Ka)1/2 can be used.
Do you know why pH = pKa in this case? At this point, the solution is a very good buffer.
Answer [OH-]=1.18e-4; pOH=4.78; pH=9.22
Justfy the following formula for this condition. [OH-]=(Kw/Ka*C)1/2
What is the pH of a 0.5 M sodium acetate solution?
Answer 0.05 M
You have doubled the volume in this case, and the concentration is 0.05 M sodium acetate.