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## AP®︎/College Chemistry

# Methods for preparing buffers

In this video, we'll explore two common methods for preparing buffer solutions. In the first approach, a certain amount of a weak acid (or weak base) is neutralized with a strong base (or strong acid), forming a conjugate acid–base pair in solution. In the second approach, a weak acid (or weak base) is combined with a salt containing its conjugate base (or conjugate acid). Created by Jay.

## Want to join the conversation?

- Why was CH3COOH + OH- used instead of NaOH + H3O+ after all NaOH is a strong base while CH3COOH is a weak acid?

(Also as a confirmation, if I wanted to use the first method in the second question, I'm supposed to use NH3 + H3O+ -> NH4+ + H2O, right?)(4 votes)- Sodium hydroxide is an ionic species so it will dissociate completely in water into sodium cations and hydroxide anions. The sodium cation doesn't contribute anything to the acid/base reaction so it is omitted. Rather it is the hydroxide anion which is acting as the strong base which we care about and therefore include solely.

Acetic acid is a weak acid meaning that most of it will be in its protonated, or neutral, form instead of being represented as hydronium ions.

For the ammonia/ammonium case you have two solutions being mixed which have significant amounts of both. So you'll have a situation where they both react with the neutral water.

NH3 + H2O ⇌ NH4+ + OH-

NH4+ + H2O ⇌ NH3 + H3O+

They're both creating acidic/basic species together which is what a buffer is.

Hope that helps.(4 votes)

- Hope you can reply asap with my question!

In8:03

How did you get the numerator (.80 mol) and denominator of (.10 mol).

Kindly explain in detail.

Additionally, when solving this type of solution is pKa always given?(1 vote)- For the Henderson-Hasselbach equation, which is the one shown at that time, we’re including the ratio of the conjugate base to the acid’s molarity with constitutes the buffer solution. With the molarity of ammonia, the base, as the numerator and the molarity of ammonium, the acid, as the denominator.

To make the buffer solution we combined two solutions of the base and acid with their original molarities and volumes known. The way Jay can skip the usual calculations and already know the final concentrations is by recognizing that the final is twice that the volume of the original solutions. And since molarity is the ratio of the number of moles of a solute by the volume of the solution, if you double to the volume, double the numerator, you half the overall ratio value. Hence the final molarities for both base and acid are half of their original values; 0.16 M/2 = 0.080 M (not 0.8M) and 0.20 M/2 = 0.10 M.

For buffer problems involving the Henderson-Hasselbach equation, in order to find the pH of the buffer the concentrations of the acid and base, and the pKa must be known. If they don’t give the pKa directly, they might instead give just the Ka or the Kb from which the pKa can be calculated. Or they might just tell you the identity of the acid/base pair and you can look up their pKa values in most chemistry textbooks.

Hope that helps.(8 votes)

- At08:06, how come the Henderson-Hasselbach equation is (weak)/(conjugate)?

I thought that the Henderson-Hasselbach equation is (conjugate)/(weak). Thanks(2 votes)- Ammonia (NH3) and ammonium (NH4^+) are both weak (base and acid respectively) and conjugates of each other. Your use of the Henderson–Hasselbalch equation is redundant since we only use this equation for buffers where we have a significant amount of a weak-conjugate acid/base pair. So they're both going to be weak conjugates with respect to each other.

A better way to think of the Henderson–Hasselbalch equation is that it is the ratio of the base over the acid, or pH = pKa + log([Base]/[Acid]).

Hope that helps.(2 votes)

- Why didn't you use the same process as you did for the first example for the second example? For the first one you converted the values to moles and made a table with that but why didn't you do that for the second example? Can you do that same process for the second example or is it not necessary?(1 vote)
- The fundamental part of creating a buffer is combining an appreciable amount of a weak acid along with its conjugate base.

The first method has the weak acid, acetic acid, but not its conjugate base, acetate. So we need a smaller amount (smaller number of moles) of hydroxide to react with some of the acetic acid to create acetate. Once all the hydroxide is reacted and we have a mixture of acetic acid and acetate, our buffer is created.

The second method simply adds both the weak base (ammonia) and its conjugate acid (ammonium) together to create the buffer. Ammonium is added as a salt since its charged, but dissociates into the appropriate ion in water.

Both methods accomplish the same task, but in different ways. Calculating the pH of a buffer solution requires the Henderson-Hasselbalch equation and knowledge of the concentrations (in molarity) of both the weak acid and its conjugate base. The second method directly introduces both the acid/base pair species and simply requires a calculation to determine the molarities of both after mixing.

The first one requires us to know how many moles of hydroxide reacted with the moles of acetic acid to produces moles of acetate. We don't really need an ICE table since we're dealing with moles and not molarities there. We know from the chemical formula that acetic acid reacts with hydroxide in a 1:1 mole-to-mole ratio, and produces acetate in the same 1:1 ratio. Once we have the moles of remaining acetic acid and acetate, we simply find the molarities of both and find the pH like the second method.

So the second method has both species already for the buffer, the first method needs to produce the conjugate base first before we do the same calculation as in the second method.

Hope that helps.(3 votes)

- on the second example (6:51) it is a weak base and its conjugate acid, am i correct? so the pKa of 9.15 is the pKa of the conjugate acid (NH4+) right? sorry if it is a stupid question but why don't we change the equation to pKb?(1 vote)
- Yes, a buffer is always a weak acid-base pair. They both have to be weak so they can coexist in solution with appreciable concentrations.

Yes, the ammonium, NH4^(+), is the acid so the pKa applies to it and not the base ammonia, NH3.

There is a version of the Henderson-Hasselbalch equation which uses pKb instead of pKa. The derivation is similar to the acid version, but it solves for pOH which is less commonly used compared to pH. It is: pOH = pKb + log((acid)/(base))

Hope that helps.(1 vote)

- what if i were to mix a weak acid and base like sodium acetate and acetic acid? how would i calculate the initial concentration of the acetate?(1 vote)
- Only a minor amount of the acid molecules of a weak acid ionizes themselves and participate in an acid-base reaction. A 1M solution of acetic acid for example only has a percent ionization of 0.42%, meaning only 0.42% of the acetic molecules turn into acetate. This percent ionization is lowered even further by the addition of acetate ions because of the common ion effect. So what this means is that the concentrations of a weak acid and its conjugate base at equilibrium in a buffer are nearly identical to their initial concentrations. So, the initial concentration of a buffer before mixing is practically the same as the concentration after mixing.

Hope that helps.(1 vote)

- Can we solve it like this?

0.16M * 0.1L = 0.016mol A^-

0.20M * 0.1L = 0.020mol Ha

0.016mol/0.020mol = 0.8mol

pH = 9.25 + Log(0.8mol)

pH = 9.15(1 vote) - For the second example, do we not let the conjugate acid and the weak base shift to equilibrium before we plug the concentrations in the henderson-hasselbalch equation? What is the difference between using the ICE table and this equation? Is the second example already at equilibrium?(1 vote)
- The Henderson-Hasselbalch equation assumes that a buffer solution has already reached equilibrium. The derivation of the equation stems from the acid-dissociation equilibrium reaction.

If we have a general acid-dissociation reaction: HA ⇌ A^(-) + H^(+)

Then the equilibrium expression becomes: Ka = ([A^(-)][H^(+)])/[HA]

Ka = [H^(+)]*([A^(-)]/ [HA]), rewritten with the H^(+) infront.

log(Ka) = log([H^(+)]*([A^(-)]/ [HA])), apply a logarithm to both sides.

log(Ka) = log([H^(+)]) + log([A^(-)]/ [HA]), a product with a logarithm can be turned into a sum of logarithms which is a property of logarithms.

-log(Ka) = -log([H^(+)]) - log([A^(-)]/ [HA]), multiply both sides of the equation by -1.

pKa = pH - log([A^(-)]/ [HA]), the negative logarithm of Ka and [H^(+)] is by definition the pKa and pH.

pH = pKa + log([A^(-)]/ [HA]), adding the logarithm of the ratio of the base and acid gives the usual representation of the Henderson-Hasselbalch equation.

So, if we use the Henderson-Hasselbalch equation, the reaction is already at equilibrium. Using an ICE table would also give you the same pH because both methods are assuming the equation reaches equilibrium.

Hope that helps.(1 vote)

## Video 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.