General Chemistry

Titration is a technique for determining a solution’s molarity. Titrations can be based on acid-base, precipitation, or oxidation-reduction reactions. In this post, we will discuss acid-base titrations which is an acid-base reaction using solutions with unknown and known standard concentrations called a standard solution.

The idea is to add just enough of the standard solution to completely react with the solution of unknown concentration.

The unknown solution is added dropwise using a buret to the standard solution. When the concentrations of H+ and OH ions are equal, the equivalence point is reached. However, 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 264 mL of a NaOH solution is needed to neutralize 1.53 g of KHP. What is the molarity of the NaOH solution?

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

The reaction can be written as:

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

${\rm{1)}}\;{\rm{n}}\;\left( {{\rm{KHP}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.53}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{204}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00749}}\;{\rm{mol}}$

2) Because the molar ratio is 1:1, the moles of NaOH reacted are also 0.00749 mol.

${\rm{3)}}\;{\rm{M (NaOH)}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00749}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.264}}\;{\rm{L}}}}\; = {\rm{0}}{\rm{.0283}}\;{\rm{mol/L}}$

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:

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

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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?

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?

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?

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.

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:

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