General Chemistry

Titration is a common laboratory technique for determining the concentration of a solute. The most common types of titrations are acid-base titrations and redox titrations. Today’s topic is the acid-base titrations so let’s see how they work.

Before the titration begins, the solution with the known concentration, also known as the titrant, is added to the burette, and the solution with an unknown concentration, the analyte, is placed in an Erlenmeyer flask under the burette. The titrant then is added dropwise to the analyte until the acid or the base in it is neutralized.

 

 

So, how do we know when it is neutralized and why do we stop at that point?

First, remember that the purpose of the titration is to demine the concentration of the unknown solution. If you wonder about it, yes, it is known as we know what is in the solution, but we don’t know the concentration, and for simplicity, we call it an unknown solution.

And the idea is this: we know the volume of the unknown solution, so if we knew the moles of the acid or the base in it, we could calculate the concentration using the formula for molarity. In general, determining the concentration or amount of the analyte by measuring volume is called volumetric analysis.

Now, the moles of the acid or the base is determined based on the stoichiometry of the acid-base reaction between the components of the standard and unknown solutions.

For example, if we want to determine the concentration of an HCl solution, we can use a standard solution of NaOH, and determine the moles of HCl using their 1 : 1 ratio:

2NaOH(aq) + HCl (aq) → NaCl(aq) + 2H2O(l)

 

The next question is, how do we know when to stop the titration? This should be done when there is no excess of an acid or a base left in the solution. In other words, the titration should be stopped when the concentrations of H+ and OH ions are equal. This is the equivalence point, and it is essentially what we have in water as very little of it is dissociated and the con concentrations of H+ and OH ions are equal (Kw – Chem II material).

To spot the equivalence point, we use an indicator which is a dye that changes its color when switching from acidic to a basic solution.

In practice though, the addition is continued slowly until the indicator signals a permanent color change, which is called the endpoint. Ideally, just one drop of addition should switch from the equivalence to the endpoint. 

 

 

This tiny amount over the equivalence point does not affect the accuracy very much and is needed to change the solution from acidic to basic which is the reason for the color change of the indicator. A common indicator used in acid-base titrations is phenolphthalein, which is colorless in acid media, and purple when the solution becomes basic.

For example, in a titration experiment, it was determined that 45 mL of a 0.10 M NaOH solution was needed to neutralize a 20.0 mL solution of KHP. What is the molarity of the KHP solution?

KHP is the abbreviation of potassium hydrogen phthalate (KHC8H4O4) which is a monoprotic acid commonly used in acid-base titration.

The reaction can be written as:

 

 

The plan is to calculate the moles of NaOH and using the stoichiometric ratio, determine the moles of KHP. Once the moles of KHP are known, we can use it to calculate its concentration using the formula for molarity:

                       

                            

\[{\rm{1) n (NaOH)}}\;{\rm{ = }}\;0.045\,\cancel{{\rm{L}}}\, \times \,\frac{{{\rm{0}}{\rm{.10}}\;{\rm{mol}}}}{{{\rm{1}}\;\cancel{{\rm{L}}}}}\; = {\rm{0}}{\rm{.0045}}\;{\rm{mol}}\]

                                                                                                       

2) Because the molar ratio is 1:1, the moles of KHP reacted are also 0.0045 mol.

 

\[{\rm{3)}}\;{\rm{M (KHP)}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0045}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0200}}\;{\rm{L}}}}\; = {\rm{0}}{\rm{.225}}\;{\rm{mol/L}}\]

 

The neutralization reaction between NaOH and KHP simplifies the calculations because of the 1:1 molar ratio. Let’s do another example, where a diprotic acid H2SO4 is used instead of KHP.

 

How many milliliters (mL) of a 0.85 M NaOH solution are needed to neutralize 135 mL of a 0.260 M H2SO4 solution?

First, write the acid-base reaction:

2NaOH(aq) + H2SO4 (aq) → Na2SO4 (aq) + 2H2O(l)

The plan is still the same, and we need to determine the moles of H2SO4 first. Make sure to convert the mL to L because the units of molarity are mol/L.

                                                    

\[\begin{array}{l}{\rm{V(}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{135}}\;{\rm{mL}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.135}}\;{\rm{L}}\\\end{array}\]

 

\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{M}}\,{\rm{ \times }}\,{\rm{V}}\;{\rm{ = }}\,{\rm{0}}{\rm{.260}}\;\frac{{{\rm{mol}}}}{{\rm{L}}}\,{\rm{ \times }}\;{\rm{0}}{\rm{.135}}\;{\rm{L}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0351}}\;{\rm{mol}}\]

 

Next, use the molar ratio to determine the moles of NaOH:
                                     

\[{\rm{n}}\;{\rm{(NaOH)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0351}}\;\cancel{{{\rm{mol}}\,{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\;{\rm{mol}}\,{\rm{NaOH}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0702}}\,{\rm{mol}}\]

 

And finally, the volume of NaOH solution is determined using the molarity formula:

 

\[{\rm{3)}}\;{\rm{V}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0702}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.85}}\;{\rm{mol/L}}}}\;{\rm{ = 0}}{\rm{.0826}}\;{\rm{L}}\;{\rm{ = }}\;{\rm{82}}{\rm{.6}}\;{\rm{mL}}\]

 

There is a lot more to acid-base titrations, in particular, the pH change during the experiment demonstrated by titration curves. These are usually covered in chem 2 curriculum, but I will leave the links here in case you need them:

 

 

Check Also

Practice

1.

27.0 mL of a NaOH solution was needed to neutralize 0.326 g of KHP (KHC8H4O4). What is the molarity of the NaOH solution?

answer

0.0593 mol/L

Solution

First, let’s write a balanced chemical equation for this reaction:

NaOH + KHC8H4O4 → NaKC8H4O4 + H2O

The plan here would be calculating the moles of KHP, using it to determine the moles of NaOH based on the stoichiometric ration, and use that to calculate the volume of NaOH solution:

 

\[{\rm{n}}\left( {{\rm{KHP}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.326}}\;{\rm{g}}}}{{{\rm{204}}{\rm{.2}}\;{\rm{g/mol}}}}{\rm{ = }}\;{\rm{0}}{\rm{.00160}}\;{\rm{mol}}\]

 

Because the ratio of NaOH and KHP is 1:1, the moles are going to be equal and therefore, there are 0.00160 mol NaOH reacting with the KHP.

Now, we can use the moles and the volume of the solution to determine the molarity of NaOH. Don’t forget to convert mL to L when working with molarity:

\[{\rm{V}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;27.0\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0270}}\;{\rm{L}}\]

 

\[{\rm{M}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00160}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0270}}\;{\rm{L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0593}}\;{\rm{mol/L}}\]

 

2.

What volume of 0.150 M calcium hydroxide is required to neutralize 250.0 mL of 0.0350 M sulfuric acid?

answer

0.0583 L

Solution

First, let’s write a balanced chemical equation for this reaction:

Ca(OH)2(s) + H2SO4(aq) → CaSO4(aq) + 2H2O(l)

The linkage between the two reactants/solutions is still going to be the mole conversion based on the stoichiometric ratio. So, to plan is as follows:

 

 

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

 

\[{\rm{V}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{250}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.2500}}\;{\rm{L}}\]

 

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]

n = MV

\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0350}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.2500}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;{\rm{mol}}\]

 

Now, we use this to determine the moles of Ca(OH)2 based on the stoichiometric ratio:

 

\[{\rm{n}}\left( {{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;{\rm{mol}}\]

 

Next, use the moles in the formula for molarity and determine the volume of the solution in L:

 

\[{\rm{V}}\left( {{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00875}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.150}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0583}}\;{\rm{L}}\]

3.

How many mL of 0.250 M HCl will neutralize 60.00 mL of 0.060 M Ba(OH)2?

answer

28.8 mL

Solution

First, let’s write a balanced chemical equation for this reaction:

Ba(OH)2(s) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)

The linkage between the two reactants/solutions is still going to be the mole conversion based on the stoichiometric ratio. So, the plan is as follows:

 

 

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

 

\[{\rm{V}}\left( {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{60}}{\rm{.00}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.06000}}\;{\rm{L}}\]

 

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]

n = MV

\[{\rm{n}}\left( {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.060}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.06000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0036}}\;{\rm{mol}}\]

 

Now, we use this to determine the moles of HCl based on the stoichiometric ratio:

 

\[{\rm{n}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0036}}\;\cancel{{{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{HCl}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0072}}\;{\rm{mol}}\]

 

Next, use the moles in the formula for molarity and determine the volume of the solution in L and convert it to mL:

 

\[{\rm{V}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0072}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.250}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0288}}\;{\rm{L}}\]

 

To get the mL, we just need to multiply by 1000, or use the molar conversion:

\[{\rm{V}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0288}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{28}}{\rm{.8}}\;{\rm{mL}}\]

4.

A 4.598-g sample of acetylsalicylic acid (the active ingredient in aspirin) was dissolved in water. It took 35.6 mL of a 0.717 M NaOH solution to neutralize the acid. Determine the molar mass of the acid considering that it is a monoprotic acid.

answer

180.3 g/mol

Solution

The plan here would be to find the moles of acetylsalicylic acid based on the amount of NaOH, and use the moles together with the given mass (4.598 g) to determine the molar mass:

 

 

We don’t know the formula of acetylsalicylic acid, but we know that it is a monoprotic acid and therefore, the stoichiometric ratio with NaOH is going to be 1:1. To find the moles of NaOH, we are going to use the formula for molarity so, first, express the volume in L:

 

\[{\rm{V}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;{\rm{35}}{\rm{.6}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0356}}\;{\rm{L}}\]

\[{\rm{n(NaOH)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.717}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0356}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0255}}\;{\rm{mol}}\]

 

Because the moles of the acid and NaOH are equal, 0.0255 mol of acid must’ve reacted. And at this point we have the moles and the mass of the acid. Use these to determine the molar mass:

 

\[{\rm{M}}\left( {{\rm{acid}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{n}}}\;{\rm{ = }}\;\frac{{{\rm{4}}{\rm{.598}}\;{\rm{g}}}}{{{\rm{0}}{\rm{.0255}}\;{\rm{mol}}}}{\rm{ = }}\;{\rm{180}}{\rm{.3}}\;g/{\rm{mol}}\]

 

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