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Chemistry LibreTexts

Titration

Skills to Develop

  • Calculate the pH and plot it during a titration of a strong acid by a strong base.
  • Calculate the pH and plot it when a weak acid is titrated by a strong base.

Titrations

The process of obtaining quantitative information of a sample using a fast chemical reaction by reacting with a certain volume of reactant whose concentration is known is called titration. When an acid-base reaction is used, the process is called acid-base titration. When a redox reaction is used, the process is called a redox titration. Titration is also called volumetric analysis, which is a type of quantitative chemical analysis.

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.

An example of titration is the acetic acid and NaOH - strong base and weak acid - titration following the equation below. 

HC2H4O2(aq)+OH-(aq) → C2H4O2(aq)+H2O(l) 

Figure 1: Macroscopic view of NaOH and acetic acid titration                               Figure 2: Submicroscopic view of the titration reaction

naoh and acetic acid titration.pngsodium and acetic acid.png

 

Titration Curve

The plot of pH as a function of titrant added is called a titration curve. Let us take a look at a titration process:

Base added \(\ce{[H+]}\) pH
0 1.0 0.0
1.0 9/11 0.087
2.0 8/12 0.176
5.0 5/15 0.477
Half- equivalence point
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
10 \(\ce{H2O\rightleftharpoons H+ + OH-}\) pH = 7
\(\ce{NaCl}\) neutral salt
Base added \(\ce{[OH- ]}\) pH
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

Example 1

Evaluate \(\ce{[H+]}\) and pH in the titration of 10.0 mL 1.0 M \(\ce{HCl}\) solution with 1.0 M \(\ce{NaOH}\) solution, and plot the titration curve.

Solution
\(\begin{align}
\textrm{The amount of acid present} &= \mathrm{V_a\times C_a}\\
    &= \mathrm{10.0\: mL \times \dfrac{1.0\: mol}{1000\: mL}}\\
    &= \textrm{10 mmol (mili-mole)}
\end{align}\)

\(\textrm{The amount of base NaOH added} = \mathrm{V_b\times C_b}\)

\(\textrm{The amount of acid left} = \mathrm{V_a\times C_a - V_b\times C_b}\)

\(\textrm{The concentration of acid and thus }\ce{[H+]}
    = \mathrm{\dfrac{[V_a\times C_a - V_b\times C_b]}{V_a + V_b}}\)

With the above formulation, we can build 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 equivalence point, why is pH=7? What formula is used to calculate pH?

Why does pH change rapidly at the equivalence 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 \(\ce{[Na+]}\) and \(\ce{[Cl- ]}\) at the following points: initially (before any base is added), half-equivalence point; equivalence point, after 10.5 mL \(\ce{NaOH}\) is added, after 20.0 mL \(\ce{NaOH}\) 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 a large amount before observing much pH change, but when the titration is at its end or equivalence point, you would like to add the titrant slowly. In the simulation, you control the rate of titration.

While trying the Java simulation of titration, please answer the following questions.

  • Identify the buffer area on the titration curve.
  • In the titration of weak acids, why does the pH change quickly at the beginning?

Questions

  1. What are the pH of solutions of 10, 1.0, 0.10, 0.010 and 0.0010 M \(\ce{HCl}\)?
  2. What are \(\ce{[H+]}\) and the pH at the half equivalence point when a solution of 1.0 M \(\ce{HCl}\) is titrated by a 1.0 \(\ce{NaOH}\) solution?
  3. What are \(\ce{[Na+]}\) at the half equivalence point when a solution of 1.0 M \(\ce{HCl}\) is titrated by a 1.0 \(\ce{NaOH}\) solution?
  4. What are \(\ce{[H+]}\) and the pH at the equivalence point when a solution of 1.0 M \(\ce{HCl}\) is titrated by a 1.0 \(\ce{NaOH}\) solution?
  5. What are \(\ce{[Na+]}\) and \(\ce{[Cl- ]}\) at the equivalence point when a solution of 1.0 M \(\ce{HCl}\) is titrated by a 1.0 \(\ce{NaOH}\) solution?
  6. What are \(\ce{[H+]}\) and the pH of 1.0 M acetic acid solution? Ka = 1.8e-5
  7. What is the pH of the above solution when half of the acid is neutralized by \(\ce{NaOH}\) in the titration?
  8. What is the pH of the end point or equivalence point when 1.0 M acetic acid is titrated by 1.0 M \(\ce{NaOH}\) solution?
  9. A 0.10 M acetic acid solution is titrated with a 0.10 M \(\ce{NaOH}\) solution to the equivalence point. What is the concentration of the acetate ion?

Solutions

  1. Answer pH = -1, 0, 1, 2, and 3
    Consider...
    No calculators should be used.

  2. Answer \(\mathrm{[H^+] = 0.333}\); pH = 0.477
    Consider...
    Do not forget the dilution factor.

  3. Answer \(\mathrm{[H^+] = 0.333}\);
    Consider...
    What about \(\ce{[Cl- ]}\)?

  4. Answer \(\ce{[H+]} = \textrm{1e-7}\); pH = 7
    Consider...
    This is only a theoretical value.

  5. Answer \(\ce{[Ca+]} = \ce{[Cl- ]} = 0.5\)
    Consider...
    Note that \(\ce{[H+]}\) is balanced by \(\ce{[OH- ]}\), amd \(\ce{[Na+]}\) is balanced by \(\ce{[Cl- ]}\).
    Most students usually forget the salt resulting from the titration.

  6. Answer \(\ce{[H+]}=0.00424\); pH = 2.37
    Consider...
    The approximation \(\ce{[H+]}=(C K_{\ce a})^{1/2}\) can be used.

  7. Answer pH = 4.74
    Consider...
    Do you know why pH = pKa in this case? At this point, the solution is a very good buffer.

  8. Answer \(\ce{[OH- ]}=1.18\textrm{e-}4\); pOH = 4.78; pH = 9.22
    Consider...
    Justify the following formula for this condition. \(\ce{[OH- ]}=(K_{\ce w}/K_{\ce a}\times C)^{1/2}\)

    What is the pH of a 0.5 M sodium acetate solution?

  9. Answer 0.05 M
    Consider...
    You have doubled the volume in this case, and the concentration is 0.05 M sodium acetate.

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