Strong acid-strong base titration
Definitions & Key takeaways
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.
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 titran 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 HC turns into H plus ions and cl minus ions.
Therefore, if the concentration of HC is 0.500 that's also the concentration of H plus ions and cl minus ions in solution.
1 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 mL 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 concentration of H plus ions and CL minus are equal, the diagram has two particles of each ion. Next, we open the stopcock on the buret 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 N AO H turns into N A 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.
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 10.00 mL 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 mL or 0.01000 L 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 iro 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 mL or 0.0200 L 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 we haven't formed any water yet.
So we put a zero under H2O. Now we wanna 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 1 to 1 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 for H plus ions.
Finally, the balanced equation shows a 1 to 1 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 get a plus 0.00500 moles in the zero 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.
Since we started with 20.00 mL of hydrochloric acid solution and we added 10.00 mL of sodium hydroxide solution. The total volume is now 30.00 mL or 0.03000 L to find the concentration of H plus ions.
We take the final number of moles of H plus ions which was 0.0050 and divide by the total volume of the solution which is 0.03000 L, 0.0050 divided by 0.03000 gives a concentration of H plus ions of 0.17 molar.
To find the ph, we take the negative log of the concentration of H plus ions which gives us a ph of 0.77. So the next point on the titration curve is at 10.00 mL of sodium hydroxide added on the X axis and a ph of 0.77.
On the y axis. We can draw in a smooth line to connect the two points on the titration curve.
Notice how the ph of the solution has increased. Let's also think about the particulate diagram at this point of the titration because the sodium hydroxide that was added neutralized half of the total acid that we had present in our solution, half of the acid should remain.
So if we add one N A plus and +10 minus, the +10 minus will react with one of the H plus ions to form water, water is not shown in the particulate diagram for clarity.
So only one H plus ion remains in the solution. We also still have the two chloride ions that we started with and the one N A plus ion that was added.
Now we continue the titration by adding another 10 mL of base to our solution. Since we've already added 10 mL, adding another 10 mL means that we've added a total of 20.00 mL of base.
Since we started with 20.00 mL of hydrochloric acid, the total volume in our flask will be 40.00 mL, lets first figure out how many moles of hydroxide ions we have added to the solution.
The base has a concentration of 0.500 molar. And we've added a total of 20.00 mL or 0.02000 L.
So using the definition of molarity, we multiply 0.500 times 0.02000 to get a total of 0.0100 moles of hydroxide ions. Next, we need to know the initial number of moles of H plus ions in the solution.
We've already calculated that there were 0.0100 moles of H plus ions present. So at this point in the titration, we have added the same number of moles of hydroxide ions as the number of moles of H plus ions that were originally present because the moles of base are equal to the moles of acid.
We've reached the equivalence point since H plus and oh minus react in a 1 to 1 mole ratio to form water. All of the H plus and oh minus ions are used up which leaves only water because the ph of water is 7.00.
The ph of the solution is now 7.00. So for a strong acid, strong base titration, the ph at the equivalence point is 7.00 we can now add this point to our titration curve.
So after 20.00 mL of base added, the PH is equal to 7.00. And we've reached our equivalence point.
If we were to graph all of the points in between this point. And the previous one, we would see that the slope of the line increases dramatically as we get closer to the equivalence point.
Looking at the particulate diagram, we've added one more sodium ion and one more hydroxide ion. The hydroxide ion reacts with the H plus ion present to form water.
Since the hydroxide ion reacts with the last H plus ion, there are no more hydroxide ions or H plus ions present. However, we still have the two chloride ions that we started with and the two sodium ions that came from adding sodium hydroxide.
So at the equivalence point, we have an aqueous solution of sodium chloride. Now let's continue our titration this time, we add only a very small amount of base 0.20 mL, which brings us to a total of 20.20 mL of base added, even though we have added only a very small amount of base, because there are no H plus ions to react with the oh minus, the PH increases dramatically to 11.4.
Going back to the titration curve, we find the point corresponding to 20.20 mL on the X axis and a ph of 11.4 on the Y axis So, even though we're just past the equivalence point, the PH has risen dramatically.
If we were to graft the rest of the titration curve, the PH continues to increase but eventually levels off. For the particulate diagram, we can add in another sodium ion and a hydroxide ion.
Since there are no H plus ions present to react with the hydroxide ion, we are past the equivalence point of the titration.
All right is a quick recap for the titration of a strong acid with a strong base. Initially, only the acid is present as 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.
The titration curve shows us that the ph changes dramatically around the equivalence point for a strong acid strong base titration.
The equivalence point occurs at a ph of 7.00.
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