About this video
This video shows two ways to prepare a buffer. Method 1 adds a strong base to a weak acid: NaOH neutralizes half the acetic acid, leaving equal moles of acetic acid and acetate so the pH equals the pKa (4.74). Method 2 mixes a weak base with a salt of its conjugate acid (ammonia with ammonium chloride), and the Henderson-Hasselbalch equation gives a pH of 9.15.
You'll be able to
- find moles of acid and base and use an ICF table to see what is left.
- explain how adding a strong base to a weak acid creates a buffer.
- prepare a buffer by mixing a weak base with its conjugate acid.
- calculate buffer pH with the Henderson-Hasselbalch equation.
Key words
- ICF table
- An Initial-Change-Final mole table for a reaction that goes to completion.
- limiting reactant
- The reactant that runs out, here the hydroxide, which neutralizes part of the acid.
- conjugate acid-base pair
- A general buffer is a weak conjugate acid-base pair, such as NH4+/NH3.
Transcript
- [Instructor] Let's look at two different methods for preparing buffer solutions. In the first method, we're gonna add an aqueous solution of a strong base, sodium hydroxide, to an aqueous solution of a weak acid, acetic acid. Our goal is to calculate the pH of the buffer solution that forms when we mix these two aqueous solutions together.
Our first step is to figure out how many moles of acetic acid that we have. So if we have 100 milliliters of a 1.00 molar solution of acetic acid, we can use the equation: molarity is equal to moles divided by liters to figure out the moles of acetic acid. Since the concentration is 1.00 molar and the volume is 100 milliliters, which is equal to 0.100 liters, x is equal to 0.100 moles of acetic acid.
We can do a similar calculation to determine the moles of strong base. So for our aqueous solution of sodium hydroxide, we have 50 milliliters of it at a concentration of 1.00 molar. So we plug in the concentration 1.00 molar into the equation for molarity, and we plug in the volume, and 50 milliliters is equal to 0.050 liters.
So solving for x, we find the x is equal to 0.050 moles of sodium hydroxide. Since sodium hydroxide is a strong base, it dissociates completely in solution. Therefore, if we have 0.050 moles of sodium hydroxide, we also have 0.050 moles of sodium cations in solution and also hydroxide anions in solution.
When these two aqueous solutions are mixed, we're mixing 100 milliliters with 50 milliliters for a total volume of 150 milliliters. Let me just go ahead and write that down here really quickly. And when the two solutions are mixed, the acetic acid will react with hydroxide anions to form the acetate anion and water.
To figure out what's left over after the reaction goes to completion, we're gonna use an ICF table, where I stands for initial, C is for change, and F is for final. We've already calculated the initial number of moles of acetic acid is equal to 0.100, and the initial number of moles of hydroxide anions is equal to 0.050.
And if we assume the reaction hasn't happened yet, the initial number of moles of the acetate anion would be zero. For this reaction, the hydroxide anion is the limiting reactant. And therefore, we're gonna use it all up in the reaction. So we write -0.050 under hydroxide in our ICF table.
Looking at the balanced equation, the mole ratio of a acetic acid to hydroxide anion is one-to-one. Therefore, if we are losing 0.050 moles of hydroxide anions, we're also losing 0.050 moles of acetic acid. So when the reaction goes to completion, all of the hydroxide anions have been used up.
Therefore, we have zero moles of hydroxide anions left over. For a acetic acid, if we started with 0.100 and we're losing 0.050, half of the acetic acid has been neutralized by the hydroxide anions, and we're left with 0.050 moles when the reaction goes to completion.
For the acetate anion, the coefficient in the balanced equation is a one. Therefore, if we're losing 0.050 on the left side of the equation, we're gonna be gaining 0.050 on the right side. So when the reaction goes to completion, we have 0.050 moles of the acetate anion.
A buffer solution consists of significant amounts of a weak acid and its conjugate base. Acetic acid is a weak acid and its conjugate base is the acetate anion. Therefore, the addition of the strong base, hydroxide, which neutralized half of the acetic acid created a buffer solution because we have significant amounts of both acetic acid and its conjugate base, the acetate anion, in solution.
Remember that our goal was to calculate the pH of this buffer solution. So first, we need to calculate the concentration of acetic acid and of the acetate anion. To find the concentration of acetic acid, we take the moles of acetic acid which is equal to 0.050, and we divide by the total volume of solution.
We already calculated when we mix the two solutions together, the total volume was 150 milliliters which is equal to 0.150 liters. So 0.050 moles divided by 0.150 liters gives the concentration of acetic acid a 0.33 molar. For the acetate anion, we also have 0.050 moles, and the total volume of solution is also 0.150 liters.
Therefore, the concentration of the acetate anion is also 0.33 molar. To find the pH of the buffer solution, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation says the pH of the buffer solution is equal to the pKa of the weak acid plus the log of the ratio of the concentration of the conjugate base divided by the concentration of the weak acid.
The weak acid present in our buffer solution is acetic acid. And at 25 degrees Celsius, the pKa value of acetic acid is equal to 4.74. The acetate anion is our conjugated base and it has a concentration of 0.33 molar, so we can plug that into the Henderson-Hasselbalch equation, and the concentration of our weak acid is also 0.33 molar.
Molars cancel and 0.33 divided by 0.33 is equal to one. And the log of one is equal to zero. Therefore, for this buffer solution, the pH is just equal to 4.74. And notice that even though I calculated the concentration of the weak acid and of the conjugate base, I didn't really have to, if you look at the Henderson-Hasselbalch equation.
Because we have concentration divided by concentration, and concentration, molarity, is moles per liter. And so it's moles divided by liters, divided by moles, divided by liters. And since both the weak acid and the conjugate base have the same total volume of solution, the total volume of solution would cancel.
And so, if we want to, we could just do a ratio of the moles and we would have gotten the same answer, a final pH of 4.74. Let's look at another method for making a buffer solution. In this case, we're gonna mix an aqueous solution of a weak base with an aqueous solution that contains the conjugate acid to the weak base.
In this example, our weak base is ammonia, NH3. The conjugate acid to ammonia is the ammonium ion, NH4+. And if we have an aqueous solution of ammonium chloride, in solution, there are ammonium ions, NH4+. So we have a weak base, NH3, and its conjugate acid, NH4+.
And when the two aqueous solutions are mixed, we'll have a significant amount of both our weak base and its conjugate acid. Therefore, we will have a buffer solution. I often say that a buffer solution consists of a weak acid and its conjugate base. In this case, I've said the buffer solution consists of a weak base and its conjugate acid.
So a more general definition for a buffer solution could be a weak conjugate acid-base pair. We can calculate the pH of the buffer solution that forms when we mix the two solutions together using the Henderson-Hasselbalch equation. In this case, our numerator is our weak base which is ammonia and our denominator is the conjugate acid to ammonia which is the ammonium ion, NH4+.
The ammonia solution had a concentration of 0.16 molar, and when we mix 100 milliliters of the ammonia solution with 100 milliliters of the solution of ammonium chloride, the total volume would double. So we went from 100 milliliters to 200 milliliters. Therefore, since we're doubling the volume, we're halving the concentration of ammonia.
So we can put in 0.080 molar for the concentration of ammonia. Ammonium chloride is a soluble salt. Therefore, if we have 0.20 molar for the initial concentration of ammonium chloride, we have 0.20 more for the initial concentration of ammonium ions. And since we're doubling the volume when we mix the two solutions together, we are halving the concentration.
So the concentration of ammonium ions in solution is 0.10 molar. So we could plug that into the Henderson-Hasselbalch equation. So here we have the concentrations plugged in, 0.080 molar is the concentration of ammonia and 0.10 molar was the concentration of the ammonium ion.
In the Henderson-Hasselbalch equation, the pKa is the pKa value of the weak acid, which is the ammonium ion, NH4+. At 25 degrees Celsius, the pKa value of the ammonium cation is equal to 9.25. Molar cancels and when we solve for the pH of the solution, the pH is equal to 9.15.
So this problem started with a weak base and a salt that contained the conjugate acid to that weak base. It's also possible to make a buffer solution starting with an aqueous solution of a weak acid and adding a salt that contains the conjugate base to that weak acid.
For example, to make another buffer, we could have started with a solution of a weak acid, acetic acid, and to that solution, we could have added something like sodium acetate. Sodium acetate is a soluble salt that dissociates completely in solution to produce sodium cations and acetate anions.
And since we would have a significant amount of both a weak acid and its conjugate base, we would have a buffer solution.
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