Strong acid-strong base titration
Video summary
An acid-base titration is a laboratory procedure used to determine the unknown concentration of a solution by neutralizing it with a known concentration of a base. It involves adding a measured amount of a standard solution, such as a base, to the unknown solution until the reaction is complete. During the titration of a strong acid, a strong base is required. As the strong base is added, the pH slowly increases. When the number of moles of acid is equal to the moles of base added, the titration has reached the equivalence point. A pH indicator is typically used to detect the endpoint of the reaction.
Transcripción
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Let's consider the titration of a strong acid, hydrochloric acid, with a strong base, sodium hydroxide.
Let's say we have 20.00 mL of a 0.500 molar solution of hydrochloric acid in a flask.
Into the burette, we place the titrant, which is a 0.500 molar solution of sodium hydroxide.
Before we start the titration, we need to calculate the initial pH of the hydrochloric acid solution.
Because hydrochloric acid is a strong acid, we assume it ionizes 100% in water which means that all the HCl turns into H plus ions and Cl minus ions.
Therefore if the concentration of HCl is 0.500, that's also the concentration of H plus ions and Cl minus ions in solution.
Since pH is equal to the negative log of the concentration of H plus ion, we take the negative log of 0.500, which gives us a pH equal to 0.301.
Titrations can be represented by titration curves.
For this titration, we put volume of sodium hydroxide added on the x axis and pH on the y axis.
Since we haven’t added any strong base yet, our first point is at 0 milliliters of base on the x axis and a pH of 0.301 on the y axis.
Titrations can also be represented by particulate diagrams.
Since we haven't added any of our base yet, the only ions present are H plus and Cl minus.
Since the concentrations of H plus ions and Cl minus are equal, the diagram has two particles of each ion.
Next, we open the stopcock on the burette and allow some of the sodium hydroxide solution to go into the flask containing the hydrochloric acid.
Let's say we add 10.00 mL of the 0.500 molar solution of sodium hydroxide.
Because sodium hydroxide is a strong base, we assume that it dissociates 100%, which means that all of the NaOH turns into an Na plus and OH minus.
Therefore, if the concentration of sodium hydroxide is 0.500, that's also the concentration of sodium ions and hydroxide ions in solution.
Let’s write an equation for the acid-base neutralization reaction that is occurring.
An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrochloric acid to form water and an aqueous solution of sodium chloride.
The net ionic equation leaves out all of the ions that don’t participate in the reaction.
These ions are called spectator ions.
Since sodium and chloride ions are present in solution both before and after the acid-base reaction, they are the spectator ions.
Therefore, we can take out the spectator ions to give the net ionic equation of OH minus plus H plus goes to H2O.
To get another data point for the titration curve, we need to calculate the pH of our solution after adding the 10.00 milliliters of base.
We can figure out how many moles of hydroxide ions we have added by using the molarity equation.
We added in 10.00 milliliters, or 0.01000 liters, of a 0.500 molar solution of sodium hydroxide.
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Transcript
Let's consider the titration of a strong acid, hydrochloric acid, with a strong base, sodium hydroxide.
Let's say we have 20.00 mL of a 0.500 molar solution of hydrochloric acid in a flask.
Into the burette, we place the titrant, which is a 0.500 molar solution of sodium hydroxide.
Before we start the titration, we need to calculate the initial pH of the hydrochloric acid solution.
Because hydrochloric acid is a strong acid, we assume it ionizes 100% in water which means that all the HCl turns into H plus ions and Cl minus ions.
Therefore if the concentration of HCl is 0.500, that's also the concentration of H plus ions and Cl minus ions in solution.
Since pH is equal to the negative log of the concentration of H plus ion, we take the negative log of 0.500, which gives us a pH equal to 0.301.
Titrations can be represented by titration curves.
For this titration, we put volume of sodium hydroxide added on the x axis and pH on the y axis.
Since we haven’t added any strong base yet, our first point is at 0 milliliters of base on the x axis and a pH of 0.301 on the y axis.
Titrations can also be represented by particulate diagrams.
Since we haven't added any of our base yet, the only ions present are H plus and Cl minus.
Since the concentrations of H plus ions and Cl minus are equal, the diagram has two particles of each ion.
Next, we open the stopcock on the burette and allow some of the sodium hydroxide solution to go into the flask containing the hydrochloric acid.
Let's say we add 10.00 mL of the 0.500 molar solution of sodium hydroxide.
Because sodium hydroxide is a strong base, we assume that it dissociates 100%, which means that all of the NaOH turns into an Na plus and OH minus.
Therefore, if the concentration of sodium hydroxide is 0.500, that's also the concentration of sodium ions and hydroxide ions in solution.
Let’s write an equation for the acid-base neutralization reaction that is occurring.
An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrochloric acid to form water and an aqueous solution of sodium chloride.
The net ionic equation leaves out all of the ions that don’t participate in the reaction.
These ions are called spectator ions.
Since sodium and chloride ions are present in solution both before and after the acid-base reaction, they are the spectator ions.
Therefore, we can take out the spectator ions to give the net ionic equation of OH minus plus H plus goes to H2O.
To get another data point for the titration curve, we need to calculate the pH of our solution after adding the 10.00 milliliters of base.
We can figure out how many moles of hydroxide ions we have added by using the molarity equation.
We added in 10.00 milliliters, or 0.01000 liters, of a 0.500 molar solution of sodium hydroxide.