Titration is simply a measurement technique, which is used to identify the concentration of a solution. This technique can also apply to gases, however for most high-school practices it will likely take place between an acid and a base. For example; you might know that the ocean water is salty (primarily NaCl), but *how salty* is it?
Examples:
How concentrated is the carbonic and phosphoric acids in Coca-Cola?
How concentrated is the acid in the Brisbane River? Is it changing over time?
This process involves identifying the type of acid/base present in the unknown solution, then choosing an appropriate base/acid to react with it, of a known concentration.
Example:
If the unknown solution contains primarily hydrochloric acid, HCl, then we would choose a simple base like NaOH to react with it for our titration.
Examples:
How concentrated is the carbonic and phosphoric acids in Coca-Cola?
How concentrated is the acid in the Brisbane River? Is it changing over time?
This process involves identifying the type of acid/base present in the unknown solution, then choosing an appropriate base/acid to react with it, of a known concentration.
Example:
If the unknown solution contains primarily hydrochloric acid, HCl, then we would choose a simple base like NaOH to react with it for our titration.
HCl + NaOH → H₂O + NaCl
1 : 1 : 1 : 1
1 : 1 : 1 : 1
Standard solution -
Aliquot - the precise volume of the unknown sample called the Analyte.
Endpoint - the pH range at which the indicator changes colour
Equivalence point - the point at which the amounts from both reactants are said to be chemically equal with respect to the mole-ratio from the chemical equation. Titre - the volume of the standard solution from the burette used.
Titration Scenario:
Let's say we've been asked to examine the acidity of the water in the river. River water is generally not known to be highly acidic, so it is likely that any acids present would be very dilute, so we would choose an appropriately dilute base to react with it. Perhaps a 0.50M solution of NaOH would do, and let's examine a 20mL river-water sample.
Standard Solution: 0.50M NaOH
Aliquot: (water sample) 20mL
After titration, the titre was found to be 5.7mL, and after analysing the water with other chemical techniques (perhaps a Mass-Spectrometer), we identify that the primary acid is likely to be hydrochloric acid: HCl.
Titre: 5.7mL = V(NaOH)
Using c = n/V, we can calculate the number of moles of NaOH used to titrate the water.
c = n / V
∴ n = cV
n(NaOH) = 0.50 mol/L x 0.0057 L
= 0.0029 mol
During the titration technique, the NaOH is continually added to the water sample until it reaches it's endpoint (colour change) which matches the equivalence point of the chemical reaction which is given by the equation below:
Aliquot - the precise volume of the unknown sample called the Analyte.
Endpoint - the pH range at which the indicator changes colour
Equivalence point - the point at which the amounts from both reactants are said to be chemically equal with respect to the mole-ratio from the chemical equation. Titre - the volume of the standard solution from the burette used.
Titration Scenario:
Let's say we've been asked to examine the acidity of the water in the river. River water is generally not known to be highly acidic, so it is likely that any acids present would be very dilute, so we would choose an appropriately dilute base to react with it. Perhaps a 0.50M solution of NaOH would do, and let's examine a 20mL river-water sample.
Standard Solution: 0.50M NaOH
Aliquot: (water sample) 20mL
After titration, the titre was found to be 5.7mL, and after analysing the water with other chemical techniques (perhaps a Mass-Spectrometer), we identify that the primary acid is likely to be hydrochloric acid: HCl.
Titre: 5.7mL = V(NaOH)
Using c = n/V, we can calculate the number of moles of NaOH used to titrate the water.
c = n / V
∴ n = cV
n(NaOH) = 0.50 mol/L x 0.0057 L
= 0.0029 mol
During the titration technique, the NaOH is continually added to the water sample until it reaches it's endpoint (colour change) which matches the equivalence point of the chemical reaction which is given by the equation below:
HCl + NaOH → H₂O + NaCl
1 : 1 : 1 : 1
1 : 1 : 1 : 1
so the ratio between NaOH and HCl is 1:1, so if we have n(NaOH) = 0.0029 mol, then at the equivalence point, there should be 0.0029 mol of HCl in the water sample.
n(NaOH)/1 = n(HCl)/1 = 0.0029 mol
If we consider the volume of the water sample, the concentration of HCl within that volume can be found:
c = n/V
∴ c(HCl) = 0.0029 mol / 0.020 L
= 0.14 mol/L
= 0.14 M of HCl
n(NaOH)/1 = n(HCl)/1 = 0.0029 mol
If we consider the volume of the water sample, the concentration of HCl within that volume can be found:
c = n/V
∴ c(HCl) = 0.0029 mol / 0.020 L
= 0.14 mol/L
= 0.14 M of HCl
