Strong acid-strong base titration

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.

Since molarity is moles over liters, 0.500 is equal to the number of moles over 0.01000.

Multiplying 0.500 by 0.01000 gives 0.00500 moles of hydroxide ions.

We are now ready to set up an ICF table.

I stands for “initial” moles, C stands for “change” in moles, and F stands for “final” moles.

We start with the “I” row, and we plug in 0.00500 moles for the initial moles of hydroxide ions.

Next we need to find the initial moles of H plus ions before any base was added.

Since we had 20.00 milliliters, or 0.0200 liters, of a 0.500 molar solution of hydrochloric acid, 0.500 is equal to the number of moles over 0.0200.

Multiplying 0.500 by 0.0200 gives 0.0100 moles of H plus ions.

Next we can plug this number into the “I” row of the ICF table.

To fill out the “I” row, we pretend that we haven’t formed any water yet.

So we put a “0” under H2O.

Now we want to fill out the “C” row of our ICF table.

Since we have many more moles of acid already present in our flask than the amount of base that gets added, we know that all of the hydroxide ions that get added will be neutralized.

So the change in hydroxide ions is equal to minus 0.00500 moles — all of the ions get consumed.

Looking at the balanced equation, we can see that there is a one-to-one mole ratio between hydroxide ions and H plus ions.

So if we lose 0.00500 moles of hydroxide, we also lose 0.00500 moles of H plus ions.

So we put minus 0.00500 in the C row for H plus ions.

Finally, the balanced equation shows a one-to-one mole ratio between hydroxide ions and water.

So for every mole of hydroxide ions that gets consumed, we produce a mole of water.

Therefore, we put a plus 0.00500 moles in the “C” row below water.

Since we start with 0.00500 moles of hydroxide and we lose 0.00500 moles, we are left with zero moles of hydroxide ions.

Since we start with 0.0100 moles of H plus ions, and we lose 0.00500 moles, we are left with 0.0050 moles of H plus ions.

To calculate the pH of the solution, we need to know the concentration of H plus ions in the solution.