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

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

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:

There is a large change in pH before and after 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

Weak acid - Strong base titration curve

Check your understanding Sketch on a piece of paper, the titration curve obtained when 20 cm³ of 0.100 mol dm^-3 ethanoic acid (pK_a = 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 = pK_a.

This is a titration of a strong base against a weak acid.

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 = pK_a 

When a solution is at pH = pK_a, 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 = pK_a, 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 = pK_a is 10 + 10 = 20cm³.

The titration curve below shows the reaction between a solution of a weak monoprotic acid, HX, and aqueous sodium hydroxide.

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

1.0 × 10⁻⁵

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