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Titration Curve Generator by Grace Leong

 

http://iwant2study.org/lookangejss/chemistry/

 

 Download model (simulator made by grace leong)

Titration Curves [H2 Chemistry]

 
What are titration curves and how can we read and interpret them?
Objectives
  1. Describe the changes in pH during acid-base titrations 
  2. Explain these changes in terms of the strengths of acids and bases 
Prerequisite knowledge
  • Concepts from mole concept and stoichiometry, theories of acids and bases, equilibria
  • Qualitative differences between strong and weak acids in terms of the extent of dissociation
  • Understanding of the terms Ka, pKa, Kb, pKb, Kw, including the relationship Kw = KaKb
  • Calculating [H+] and pH values for strong acids, weak monoprotic acids, strong bases, and weak monoacidic bases
  • Explaining how buffer solutions control pH
  • Calculating of pH of buffer solutions, given appropriate data

About Introduction to Titration Curves https://library.opal.moe.edu.sg/ictc&func=view&rid=2348

Titration is the slow addition of a solution with known concentration (called a titrant) to a known volume of another solution with unknown concentration until the end point. It is often used in medical and food industries where unknown amounts of different substances can be determined. 
You should be familiar with performing acid-base titrations in the chemistry laboratory. During a titration, the pH of the solution changes. A titration curve depicts the changes in pH of the solution as the titrant (the solution in the burette) is added to the solution in the conical flask. In short, atitration curve is a graph of pH against volume of titrant added.

Can you predict the shape of the titration curve obtained when a solution of NaOH was titrated against HCl?

taken from https://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html
It is important to learn the skill of reading a titration curve. It provides the training to read scientific graphs where different regions and different points each have their own significance. These are skills you might need if you decide to pursue science at a higher level, or engage in STEM related careers. 

In this lesson, you will be working with a titration curve generator, a simulation to generate and explore the shapes of titration curves involving strong or weak acids and bases.

If you are not yet familiar with the usage of the titration curve generator, please read the introduction tab or watch this video to learn how to use it.

Exploring pH changes using the titration curve generator 

Change the parameters and explore the features of the titration curve generator. Generate the shapes of the various titration curves

  • Strong acid - strong base

  • Strong base - strong acid

  • Weak acid - strong base

  • Strong base - weak acid

  • Weak base - strong acid

  • Strong acid - weak base

Change these parameters:

  • Concentration of solutions

  • Type of reagent - monoprotic or diprotic

  • Values of Ka (for weak acids) /Kb (for weak bases)

 

Observe what happens to the:

  • General shape of curve
  • Volume required for complete neutralisation
  • pH at point of complete neutralisation

What is common across all titration curves at the point of complete neutralisation? 

When a strong acid is changed to a weak acid of the same concentration, the volume of the same base required for complete neutralisation

stays the same

The approximate pH change at the point of complete neutralisation for a strong acid - weak base titration is from

3 to 7

Strong acid - Strong base titration curve

Do you know the explanations behind the shape of the titration curve? Can you visualise what is happening in the reaction mixture at different points on the titration curve?
Let's take a closer look at the typical points of interest on a titration curve, starting off with a strong acid - strong base titration. Pay special attention to the species present at each point of the titration.
Have your calculator and some rough paper handy to perform some pH calculations as you go through the interactive!

Weak acid - Strong base titration curve

What happens when the strong acid is replaced by a weak acid? How does the curve change and why does it change? Let's explore. As before, the important things to note are the species present at each point of the titration.
Once again, have your calculator and some rough paper handy to perform some pH calculations as you go through the interactive!

Check your understanding Sketch on a piece of paper, the titration curve obtained when 20 cm3 of 0.100 mol dm-3 ethanoic acid (pKa = 4.7) is added to 10.0 cm3 of 0.100 mol dm-3 sodium hydroxide. In your sketch, pay attention to the initial pH, pH at equivalence point ,volume of ethanoic acid required to reach equivalence point, volume of ethanoic acid required for pH = pKa.

This is a titration of a strong base against a weak acid.
TitrationQ5ans
Points (i) to (v) of the graph above are explained below:
(i) initial pH
At the beginning, only strong base is present in the solution, hence, the initial pH = pH of the strong base
pOH = -lg(0.100) = 1
pH = 14 - 1.00 = 13
 
(ii) pH change during neutralisation 
The shape of the curve reflects a sharp change in pH before and after complete neutralisation (equivalence point).
(iii) pH at equivalence point
At equivalence point, the pH is greater than 7 as a basic salt is formed from the strong base - weak acid titration. A basic salt hydrolyses in water to produce OH-. 
Equation showing hydrolysis of basic salt, CH3COO-Na+
CH3COO- + H2O http://www.w3.org/1998/Math/MathML" class="wrs_chemistry">⇌" class="Wirisformula" role="math" alt="rightwards harpoon over leftwards harpoon" style="box-sizing: border-box; min-width: auto; min-height: auto; padding: 0px; margin: 0px; outline: none; display: inline-block; vertical-align: -1px; border: none; height: 13px; width: 20px;"> CH3COOH + OH-
(iv) volume of ethanoic acid required to reach equivalence point
http://www.w3.org/1998/Math/MathML" class="wrs_chemistry">CH3COOH + NaOH→CH3COONa +H2O" class="Wirisformula" role="math" alt="CH subscript 3 COOH space plus space NaOH rightwards arrow CH subscript 3 COONa space plus straight H subscript 2 straight O" style="box-sizing: border-box; min-width: auto; min-height: auto; padding: 0px; margin: 0px; outline: none; display: inline-block; vertical-align: -4px; border: none; height: 16px; width: 317px;"> (they react in a 1:1 ratio)
Amount of sodium hydroxide = (0.100)(10/1000) = 1.00 x 10-3 mol
Amount of ethanoic acid = 1.00 x 10-3 mol
Volume of ethanoic acid = http://www.w3.org/1998/Math/MathML">1×10-30.100" class="Wirisformula" role="math" alt="fraction numerator 1 cross times 10 to the power of negative 3 end exponent over denominator 0.100 end fraction" style="box-sizing: border-box; min-width: auto; min-height: auto; padding: 0px; margin: 0px; outline: none; display: inline-block; vertical-align: -12px; border: none; height: 40px; width: 73px;"> = 10 cm3
 
For those who are interested, the following shows the steps to calculate the pH of a basic salt at equivalence point:
Since concentration of NaOH and CH3CO2H are the same and they react in 1:1 mole ratio, volume of ethanoic acid required to reach equivalence point = 10 cm3
Total volume of solution at equivalence point = 10 + 10 = 20 cm3
http://www.w3.org/1998/Math/MathML" class="wrs_chemistry">CH3COO-+H2O⇌CH3COOH+OH-" class="Wirisformula" role="math" alt="CH subscript 3 COO to the power of minus plus straight H subscript 2 straight O rightwards harpoon over leftwards harpoon CH subscript 3 COOH plus OH to the power of minus" style="box-sizing: border-box; min-width: auto; min-height: auto; padding: 0px; margin: 0px; outline: none; display: inline-block; vertical-align: -4px; border: none; height: 16px; width: 279px;">
Amount of sodium ethanoate formed = 1.00 x 10-3 mol
Concentration of sodium ethanoate = http://www.w3.org/1998/Math/MathML">1×10-3201000" class="Wirisformula" role="math" alt="fraction numerator 1 cross times 10 to the power of negative 3 end exponent over denominator begin display style bevelled 20 over 1000 end style end fraction" style="box-sizing: border-box; min-width: auto; min-height: auto; padding: 0px; margin: 0px; outline: none; display: inline-block; vertical-align: -19px; border: none; height: 47px; width: 76px;"> = 0.05 mol dm-3
http://www.w3.org/1998/Math/MathML">[OH-]=Kbc=10-1410-4.7(0.05)=4.087x10-6pOH=5.300pH = 14 - pOH=8.70 (3sf)" class="Wirisformula" role="math" alt="left square bracket OH to the power of minus right square bracket equals square root of straight K subscript straight b straight c end root equals square root of 10 to the power of negative 14 end exponent over 10 to the power of minus to the power of 4.7 end exponent left parenthesis 0.05 right parenthesis end root equals 4.087 straight x 10 to the power of negative 6 end exponent pOH equals 5.300 pH space equals space 14 space minus space pOH equals 8.70 space left parenthesis 3 sf right parenthesis" style="box-sizing: border-box; min-width: auto; min-height: auto; padding: 0px; margin: 0px; outline: none; display: inline-block; vertical-align: -82px; border: none; height: 161px; width: 219px;">
 
(v) volume of ethanoic acid required for pH = pKa 
When a solution is at pH = pKa, the buffer solution is at maximum buffering capacity. (If you have forgotten what a buffer solution is, you can re-visit the SLS lesson Introduction to Buffers)
When pH = pKa, the solution contains equal amounts of sodium ethanoate (conjugate base) and ethanoic acid (weak acid). 
Before equivalence point, every drop of ethanoic acid added will react with sodium hydroxide, forming sodium ethanoate. There will be no ethanoic acid present. 
At equivalence point, only sodium ethanoate will be present. All amounts of sodium hydroxide will also have been reacted.
After equivalence point, the sodium ethanoate will remain in the solution. In addition, since there will no longer be any sodium hydroxide left, every drop of excess ethanoic acid added will also remain in the solution as it does not react. In this titration, the buffer region will only be formed after the equivalence point. 
The first 10cm3 of ethanoic acid added formed sodium ethanoate at the equivalence point. In order for the solution to contain an equal amount of sodium ethanoate and ethanoic acid, another 10cm3 of ethanoic acid needs to be added after the equivalence point. Therefore, the volume of ethanoic acid required for pH = pKa is 10 + 10 = 20cm3.
The titration curve below shows the reaction between a solution of a weak monoprotic acid, HX, and aqueous sodium hydroxide.
TitrationQ6

What is the value of Ka, in mol dm−3, for this acid HX?

1.0 × 10−5

Good job! When 10 cm3 of NaOH was added, half the HX has reacted to form salt. There is equal amount of excess HX and X- present and pH = pKa.
Since pH = 5, pKa = 5 and Ka = 10-5

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