Buffers & The Henderson Hasselbalch Equation

Identifying a Buffer

A buffer is a solution composed of a weak acid with its conjugate base. 

Concept: Understanding a Buffer.

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Video Transcript

Hey guys! In this new video, we're going to take our first looks at what exactly is a buffer and how exactly do they work.
So we're going to say here, solutions that contain a weak acid and its conjugate base. And remember, when I say weak acid we could think of HF. Conjugate base means it has 1 less H+, so take off the H+ you have F-. But professors don't like to give you just F-, they'll combine that F- with a metal, so we could say our conjugate base here is NaF. So solutions which contain a weak acid like HF and a conjugate base has one less H like NaF are called buffer solutions because they resist drastic changes in pH.
Now how do they accomplish this? Well they do this by keeping your H+ concentration and your OH- concentration constant, it keeps them basically the same. The way this works is we're going to say adding a small amount of strong base and the pH will increase because remember if you add base you're going to go above 7, so you keep going up and up. But it's not going to increase by much because the weak acid will neutralize this strong base that you add. Now adding a small amount of strong acid and the pH will decrease. Again, it's not going to decrease by much because why? The conjugate base will neutralize what you add. So that's how a buffer works. Any acid or base that I add they get reacted with either the weak acid or the conjugate base. So remember acid and bases are natural enemies of each other. So if I had base, my weak acid steps in to try to take it out. If I add acid, my conjugate base steps up and tries to take it out. That's how a buffer will work. And remember, the only thing that can destroy a buffer is by adding too much strong base or too much strong acid. Adding water to a base does nothing to it. Well adding water to not a base but a buffer does nothing to it. Why? Because adding water would change the concentration of my weak acid but also change the concentration of my weak base proportionally, so proportionally the ratio would stay the same, so my buffer would be unaffected by the addition of water. So remember, adding a strong acid or base is the only way to really destroy a buffer, adding water does nothing to it.
Now knowing this, I want you guys to try to answer this first Practice Question. We're going to say: Which one of the following combinations does not create a buffer?
Remember we just set up above, a buffer is a weak acid and its conjugate base. So this is going to require you guys to remember the rules that we talked about in identifying compounds that is either being acidic or basic, strong or weak. Once you can do that, you will be able to guide your way to the correct answer for this one. Come back, click on the explanation button and see the choice that I make, and see if it matches up with yours.

Buffers resist drastic changes to the pH if a strong acid or strong base is added. 

If a strong base is added then the buffer resists a pH change by having the weak acid neutralize it. 

If a strong acid is added then the buffer resists a pH change by having the conjugate base neutralize it. 

Problem: Which one of the following combinations does not create a buffer?

a)     HC2H3O2 and K C2H3O2

b)     H2SO3 and NaHSO3

c)     H3PO4 and NaH2PO4

d)     HNO3 and KNO3

e)     NH4Cl and NH3

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Problem: Which of the following combinations can result in the formation of a buffer?

a)     HF  and  HI

b)     HC2H3O2   and  NH3

c)     CH3CH2NH2   and   CH3CH2NH3+

d)     NaCl and NaOH 

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Buffer Creation

As we stated earlier a buffer is composed of a weak acid and its conjugate base, but there are actually three ways to create a buffer.

Concept: Method 1 for a Creating a Buffer. 

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Video Transcript

Hey guys! In this new video, we're going to take a look at one of the different ways to create a buffer.
We know what a buffer is, a buffer is made up of a weak acid and conjugate base or a conjugate acid and a weak base, so that's what a buffer is. But there are 3 ways to make a buffer and it's going to be important that you guys recognize these 3 ways.
Now we're going to say the First Way is obvious: A buffer is made up of a weak acid and its conjugate base, so one way to make a buffer is just mixing those two things together.
So we're going to say mixing a weak acid and its conjugate base. So for Example:
What we need to realize here is that we're going to say in this case a good buffer, an ideal buffer, so ideal buffer means that this is the best type of buffer. An ideal buffer is when our weak acid = our conjugate base, they're the same amount. So let's say we have 0.1 M HF and 0.1 M NaF. And we should realize that the numbers could be different for them but they're work best when they're the same. We're going to say this has to do with our buffer range. What you should realize here is a good buffer has to fall within the range of 10:1 or 1:10. Okay, so what do I mean by that? I mean this.. So 10:1 that means our weak acid utmost could only be 10 times more than my conjugate base, if it falls outside that range it will be a bad buffer. So good buffer is between 10:1 ratio or 1:10 ratio. In the 1:10, now the weak acid is 1 and the conjugate base is 10 times that. Okay, so again this is the buffer range, a good buffer falls within this range. If it falls outside that range it'll still be a buffer, it will just be a very bad buffer. And remember, an ideal buffer is the best buffer, that's when both are the same.
Now connected to this also was another topic which is called our buffer capacity. Now we're going to say buffer capacity, all it means is the more concentrated my weak acid and conjugate base, the better my buffer. And this makes sense because remember what is the buffer doing? It's trying to neutralize the strong acid and the strong base that you add to them. And they can't do their job effectively if there's a little bit of them. So to fight such strong acids and strong bases, you need a lot of weak acid and a lot of conjugate base. The more the better, the more they can better defend themselves. So here we're going to say this is good because they're both the same number but this is better because it's more concentrated. So this has a better buffer capacity. So buffer capacity the more concentrated, the better.
Now if we go back to this ideal buffer, we're going to say that's when they both equal each other. We're going to say that this is found at our half-equivalence point. When we reach the half-equivalence point, that's when we have an ideal buffer. We'll talk more on detail on that when we get to titration graphs. When we get to the titration graphs of buffers, we'll take a close look and see what do I mean by the half-equivalence point and why is this an ideal buffer. So just for right now, we'll write a little note on that. An ideal buffer is when they're both equal, this happens at the half-equivalence point. So remember the difference between buffer capacity and buffer range. The best buffer is when they're both equal in amount. The buffer range says they could be a 10:1. Once it's outside that 10:1 ratio, the buffer is still a buffer but it's going to be a bad buffer. It'll get destroyed very quickly by a strong acid or base that we add. 

The first and most obvious way to create a buffer is to simply combine a weak acid and its conjugate base. In this case, a buffer is most ideal when both components are highly concentrated and equal to one another. 

The weak acid and conjugate base can be different from one another by up to a magnitude of 10. This is called the buffer range. If they are different by more than 10 then this is considered a bad buffer.

Concept: Method 2 for a Creating a Buffer. 

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Video Transcript

Now the Second Way to make a buffer is to mix some strong acid with some weak base. But here's the thing about this type, we're going to say here the weak base has to be higher in amount, must be higher in amount. If the weak base is equal to the strong acid, it'll destroy the buffer. If the strong acid is greater than my weak base, it'll destroy the buffer. So the weak base has to always be higher than the strong acid, otherwise the buffer will be destroyed.
So if we take a look at the Example we can say here:
We have 1 mol of HF and 1.5 mols of NH3. NH3 are weak base and there's more of it. Our units here can be mols or Molarity. In this case this would be a buffer. If we have 1 mol HF and 1 mol NH3, no longer a buffer because the strong species is equal to my weak species. If I have.. Sorry this is HCl actually. HF is a weak acid so disregard that, HCl. And if I have 1.5 mols of HCl and still 1 mol NH3, this will also no longer be a buffer. Your strong species cannot be equal to or greater than your weak species, if it is your buffer gets destroyed, so remember that. The weak species has to be higher in amount than the strong species.

The second method in creating a buffer is mixing a strong acid with a weak base. In this case since we have a strong species mixing with a weak species then we must make sure the weak species is higher in amount. 

Concept: Method 3 for a Creating a Buffer.

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Video Transcript

And then finally the last type, we just said strong acid and weak base. So, let's reverse it, weak acid and strong base. So example here, we could have 1.2, different number, 2.5 0 molar HF now, that's a weak acid and 1.0 molar NaOH, okay? So, now our weak acid is definitely larger in amount. So again, here we need the weak species higher in amount. So, this would be a buffer, got to remember that. So, if we go back first way is to mix weak acid and conjugate base here, they could be equal to or differ from each other but the good buffers are within the range of 10 to one, in the next two the weak species has to be higher than the strong species otherwise it won't be a buffer, you have to remember all three cases because you'll be tested on them and if you don't know which is which you won't be able to get the correct answer. So, just remember, weak acid conjugate base that can be equal that can be different as long as it's within the range of 10 to one it's a good buffer, if it's outside that range it's a bad buffer, case two and three the weak species have to be higher than the strong species, the strong species cannot be equal to or greater, if it is the buffer gets destroyed. Remember these principles and they'll help guide you to see if you have a buffer or not.

The third method in creating a buffer is mixing a strong base with a weak acid. In this case since we have a strong species mixing with a weak species then we must make sure the weak species is higher in amount. 

Example: Which of the following combinations can result in the formation of a buffer?

a)     0.01 moles HClO (hypochlorous acid) and 0.05 moles of NaOH.

b)     0.01 moles HClO (hypochlorous acid) and 0.05 moles of HCl.

c)     0.01 moles HClO (hypochlorous acid) and 0.05 moles of NH3.

d)     0.01 moles HClO (hypochlorous acid) and 0.001 moles of NaOH

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Video Transcript

Hey guys! In this new video, we're going to put to practice some of the principles we learned in creating a buffer.
So for Example 1 it says: Which of the following combinations can result in the formation of a buffer?
So what we should realize here is in every example we have 0.01 moles of hypochlorus acid (HClO). Now HClO it's an acid, so we know it's some type of acid. It has Oxygen so it's an oxy-acid. And remember if we do the math, 1 Oxygen - 1 Hydrogen gives us 0 left over. So HClO is a weak acid. So every single one of these is a weak acid. And because of that you have to remember what 2 ways can we make a buffer if we have a weak acid? So remember the 2 ways is if we have a weak acid and its conjugate base, and the second way is if we have a weak acid and a strong base. But in the second case, the weak species has to be higher in amount.
So let's take a look at the what's mixing with. Here we have NaOH which is a strong base. Here we have HCl which is a strong binary acid. Here NH3 is a neutral amine, so it's a weak base. And NaOH again is a strong base. So we're looking for the weak acid to be paired up with the conjugate base or the strong base. So automatically these 2 are out because we can't pair up a weak acid with a strong acid or weak acid with a weak base, it has to be a conjugate base or a strong base. Now in the first one we actually have more moles of these strong species, so this would not be a buffer, this would actually destroy the buffer. But for the last one, we have more of the weak species than the strong species, so would be the last one that makes a buffer. So again, remember the 3 principles that we learned in creating a buffer. Apply them to this type of question.
Now for Practice 1:
I want you guys to figure out which one of these combinations will result in a buffer. Here remember the word "of" means multiply. So if you multiply these 2 numbers together you can have to find moles because remember moles = Liters * Molarity so divide each of these by 1000, multiply it by the Molarity to have moles.
You could also keep it in milliliters and multiply times the Molarity but then that will give you millimoles. So I don't know if you guys want to deal with millimoles, it's basically the same thing but you don't have to do it that way. So divide them by a 1000, the mL to get L, multiply times the Molarity to give you moles.
And remember the 3 combinations that can result in the formation of a buffer. So check for each one of these, look to see do you have a weak acid and conjugate base; do you have a weak acid and strong base; or do you have a strong acid and weak base. Look for these combinations, if you don't see them don't bother doing the math. So I recommend looking at each compound first and seeing if they match up with any of these combinations. If they do, look more deeply, look to see does it help to create a buffer. If it doesn't match one of these 3 combinations, don't even bother looking at it, don't do the math at all. And remember one key thing I'll give you guys a hint, you have to find out to make sure you have the correct concentration of your strong bases. What do we say about strong bases? The concentration I give you is depended on how many ions you have of which particular for ions. Doing that will help to get the correct answer.
Once you're done with this, come back and click on the explanation button and watch me explain how to best approach this problem. Good luck guys!

Problem: Which of the following combinations can result in the formation of a buffer?

a)     50 mL of 0.10 M HF with 50 mL of 0.10 M NaOH.

b)     50 mL of 0.10 M HNO2 with 25 mL of 0.10 M Ca(OH)2.

c)     50 mL of 0.10 M CH3CO2H with 60 mL of 0.10 M NaOH.

d)     50 mL of 0.10 M HF with 30 mL of 0.10 M NaOH. 

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Problem: A buffer solution is comprised of 50.0 mL of a 0.100 M HC2H3O2 and 60.0 mL of a 0.100 M NaC2H3O2. Which of the following actions would completely destroy the buffer?

 

a) Adding 0.003 mol HC2H3O2

b) Adding 0.007 mol Ca(C2H3O2)2

c) Adding 0.005 mol NaOH

d) Adding 0.004 mol HCl

e) Adding 0.001 mol HCl

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Henderson – Hasselbalch Equation

Concept: The Henderson – Hasselbalch Equation.

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Video Transcript

Hey guys, in this new video, we’re finally get to see how do we calculate the pH of a buffer. We’re going to say we learned that whenever we had a weak acid or a base, we have to use our favorite friend, the ICE chart in order to find pH or pOH. The great thing is now that we're dealing with buffers, we no longer necessarily need to do an ICE chart. We're going to say anytime we know we have a buffer for sure, we can skip the Ice chart altogether and just use our Henderson – Hasselbalch equation. So, Henderson-Hasselbalch. It's also known as the buffer equation. It says that pH equals pKa plus the log of conjugate base over weak acid. When I say pKa, remember P just means negative log. pKa equals the negative log of Ka. Here when I’m talking about conjugate base or weak acid, the units here can be either in molarity or in moles. It all depends. If they give you only molarity, use molarity. If they give you volumes of a particular molarity, remember the word of means multiply. We’d multiply liters times molarity to give ourselves moles.

Whenever we have a buffer we can skip the ICE Chart and use the Henderson Hasselbalch. 

Example: What is the pH of a solution consisting of 2.75 M sodium phenolate (C6H5ONa) and 3.0 M phenol (C6H5OH). The Ka of phenol is 1.0 x 10-10

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Video Transcript

Let's look at this first question and see how exactly do we use it. Here it says: What is the pH of a solution consisting of 2.75 molar sodium phenolate, which is this compound, and 3.0 molar phenol, which is this compound. The Ka of phenol is 1.0 times 10 to the negative 10. What you should realize here is phenol has what? It has H's, non-metals, and oxygen. Phenol is an oxyacid. Use the Math that we've learned. You take the number of oxygens minus by the number of hydrogens. We don't get two oxygens left, so this is a weak oxyacid. We definitely know it’s weak because look at its Ka. Its Ka value is extremely small. If your Ka value is less than 1, you're a very weak acid.
Here we have a weak acid. sodium phenolate has one less H available, so this is the conjugate base. We have a weak acid, we have a conjugate base and therefore, we have a buffer. Because we have a buffer, we need to use the buffer equation, pH equals negative log of Ka plus the log of, now they only gave us the molarity here, so that's what we're going to plug in: 2.75 molar the conjugate base over 3.0 molar, our weak acid. When we plug all that in, we’ll get back the answer of 9.96

Problem: Calculate the pH of a solution formed by mixing 200 mL of a 0.400 M C2H5NH2 solution with 350 mL of a 0.450 M C2H5NH3+ solution. (Kb of C2H5NH2 is 5.6 x 10-4).

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Example: What is the buffer component concentration ratio, [Pr ] / [HPr] , of a buffer that has a pH of 5.11. (The Ka of HPr is 1.30 x 10-5). 

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Example: Over what pH range will an oxalic acid (H2C2O4) / sodium oxalate (NaHC2O4) solution work most effectively? The acid dissociation constant of oxalic acid is 6.0 x 10-2.

a) 0.22 – 2.22        b) 1.00 – 3.00        c) 0.22 – 1.22        d) 2.0 – 4.0

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Problem: PRACTICE: Determine how many grams of sodium acetate, NaCH3CO2 (MW: 82.05 g/mol), you would mix into enough 0.065 M acetic acid CH3CO2H (MW: 60.05 g/mol) to prepare 3.2 L of a buffer with a pH of 4.58. The Ka is 1.8 x 10-5.

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Example: Which weak acid-conjugate base combination would be ideal to form a buffer with a pH of 4.74.

a)     Cyanic acid and Potassium cynate           (Ka = 4.9 x 10-10)

b)     Benzoic acid and Lithium benzoate         (Ka = 6.3 x 10-5)

c)     Acetic acid and Sodium acetate                   (Ka = 1.7 x 10-5)

d)     Ammonium chloride and Ammonia         (Ka = 5.56 x 10-10)

e)     Formic acid and Cesium formate                (Ka = 1.7 x 10-4)

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Problem: A buffer solution is made by combining a weak acid with its conjugate salt. What will happen to the pH if the solution is diluted to one-fourth of its original concentration?

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Buffers & The Henderson Hasselbalch Equation Additional Practice Problems

Which pair of compounds will form a buffer in aqueous solution?

A. HCl and NaOH

B. NaCN and KCN

C. HCN and HCl

D. HCN and NaCN

E. NaCN and NaOH

F. HCl and NaCl

Watch Solution

The addition of hydrochloric acid and __________ to water produces a buffer solution.

a. HC6H5O  

b. NaOH

c. NH3

d. HNO3

e. NaNO3

Watch Solution

Which of the following solutions is a good buffer system?

A) A solution that is 0.10 M HC2H3O2 and 0.10 M LiC2H3O2

B) A solution that is 0.10 M HBr and 0.10 M KC 2H3O2

C) A solution that is 0.10 M HI and 0.10 M NH 4+

D) A solution that is 0.10 M NaOH and 0.10 M KOH

E) None of the above are buffer systems.

Watch Solution

Calculate the pH of a solution made by mixing 8.627 g of sodium butanoate in enough 0.452 M butanoic acid, HC4H7O2 , to make 250.0 mL of solution.

A. 4.75

B. 4.82

C. 5.00

D. 2.58

E. 4.65

Watch Solution

Which pair of substance can be dissolved together (in the right ratio) to prepare a buffer solution?

i, C2H3O2H
ii. NaC2H3O2
iii. NaOH
iv. HCl
v. NaCl

A. i and ii

B. i and iii

C. ii and iii

D. i and ii, ii and iv, i and iii

E. iv and v; i and iii

Watch Solution

What is the pH of a buffer solution that contains 0.55 M methylamine, CH 3NH2, and 0.29 M of methylammonium chloride, CH3NH3Cl?

A. 3.63

B. 10.32

C. 10.92

D. 10.37

E. 3.08

Watch Solution

To create a buffer that maintains a pH around 7.54, which solution would you choose?

A. CH3COOH and NaCH3COO

B. HClO and KClO

C. NaOH and HCN

D. HNO3 and KNO3

Watch Solution

Blood contains a buffer of carbonic acid (H2CO3) and hydrogen carbonate ion (HCO3-) that keeps the pH at a relatively stable 7.40. What is the ratio of [HCO3-] / [H2CO3] in blood? Ka1 = 4.30x10-7 for H2CO3

  1. 0.0926
  2. 10.8
  3. 3.98 x 10-8
  4. 1.71 x 10-14
Watch Solution

What would be the final pH if 0.0100 moles of solid NaOH were added to 100mL of a buffer solution containing 0.600 molar formic acid (ionization constant = 1.8x10-4 ) and 0.300 M sodium formate?

  1. 3.65
  2. 4.05
  3. 3.84
  4. 3.35
Watch Solution

Which of the following pairs of solution does not produce a buffer?

A. 0.3 M HClO& 0.2 M KOH

B. 0.2 M NH3 & 0.1 M HClO4

C. 1.5 M HF & 1 M LiOH

D. 0.1 M KCOOH & 0.1 M NaCOOH

Watch Solution

Which of the following procedures will result in the largest change in pH?

Adding NaNO2 to a solution of HNO2 in water
or 
Adding NaNO3 to a solution of HNO3 in water

 

Watch Solution

One liter of a buffer from the blood is found to have 0.4 moles of acid (pK = 4.76) and 0.1 moles of conjugate base.

a. Is this pH of buffer greater than, equal to, or less than 4.76? Show your reasoning mathematically?

 

b. You want to get the buffer to maximum buffering capacity. You have HCl and NaOH available. Which would you add and how much in terms of moles?


c. Is the pH of the buffer after step b greater than, equal to, or less than to 4.76?

 

 

Watch Solution

Calculate the pH of a buffer solution that contains 0.25 M chloroacetic acid (ClCH2CO2H) and 0.15 M sodium chloroacetate (ClCH2CO2Na).  [Ka = 1.4 × 10–3  for chloroacetic acid]

A)  3.07    B)  2.85    C)  4.19    D)  2.63    E)  4.41

Watch Solution

Which one of the following combinations CANNOT function as a buffer solution?

A)        H2C2O4 and CsHC2O4             D)       HF and LiF

B)        NH3 and (NH4)2SO4                 E)        HCl and NaCl

C)        K2HPO4 and KH2PO4           

 

Watch Solution

When to use the Henderson-Hasselbalch Equation

  • Anytime we have a buffer we use it.
  • Buffer  =  Weak Acid  &  Conjugate Base        OR          Conjugate Acid   &  Weak Base
  • A buffer can be destroyed by adding ____________ or ____________.

 

 

There are 3 ways to form a buffer:

  1. Mixing a WEAK acid and its CONJUGATE base.

An ideal buffer, in this case, is when both are equal in amount. An ideal buffer happens at the half-equivalence point.

EX: _______________ HF  and _______________ NaF

 

 

 

 

 

      2. Mixing a  STRONG acid and a  WEAK base.

In this case, the  WEAK base must be higher in amount than the  STRONG acid.

          EX:  _______________ HCl and  _______________ CH3NH2

 

      3. Mixing a  WEAK acid and a  STRONG base.

In this case, the  WEAK acid must be higher in amount than the  STRONG base.

              EX:    _______________ HNO2  and  _______________ NaOH

Watch Solution

Determine how many grams of potassium hypochlorite, KClO (MW: 90.55 ), you would mix into enough 0.055 M hypochlorous acid, HClO (MW: 52.46 ) to prepare 5.0 L of a buffer with a pH = 6.77. Ka of HClO is 2.9 x 10-8.

Watch Solution

Which of the following buffer systems would you use to create a buffer with a pH of 4.0? The Ka of nitrous acid, HNO2 is 4.5 x 10-4.

a) 0.30 M HNO2 / 0.22 M NaNO2

b) 0.22 M HNO2 / 0.30 M NaNO2

c) 0.11 M HNO2 / 0.50 M NaNO2

d) 0.50 M HNO2 / 0.11 M NaNO2

e) 0.30 M HNO2 / 0.30 M NaNO2

Watch Solution

Find the pH of a buffer solution made by mixing 100 mL of 0.20 M methylamine, CH3NH2, with 250.0 mL of a 0.50 M CH  3NH3Cl. Kb of methylamine is 4.4 x 10-4

 

Watch Solution

Over what pH range will an oxalic acid (H 2C2O4) / sodium oxalate (NaHC2O4) solution work most effectively? The acid dissociation constant of oxalic acid is 6.0 x 10-2.

a) 0.22 – 2.22                 b) 1.00 – 3.00                  c) 0.22 – 1.22                   d) 2.0 – 4.0

Watch Solution

Which weak acid-conjugate base combination would be the best to create a solution with a pH of 4.193?

     a)  Butanoic acid and Potassium butanoate (Ka = 1.5 x 10-5)

     b)  Benzoic acid and Sodium benzoate (Ka = 6.3 x 10-5)

     c)  Acetic acid and Lithium acetate (Ka = 1.8 x 10-5)

     d)  Ammonium chloride and Ammonia (Kb = 1.8 x 10-5)

     e)  Hypochlorous acid and sodium hypochlorite (Ka = 2.9 x 10-8)

Watch Solution

A buffer solution is made by combining a weak acid with its conjugate salt. What will happen to the pH if the solution is diluted to one-half of its original concentration?

a) The pH will increase.

b) The pH will decrease.

c) The pH will remain constant.

d) The solution will become more neutral. 

Watch Solution

Explain why the salt of a weak acid, as well as the acid itself, must be present to form a buffer solution.

1. The anion from the salt is needed to partially neutralize added base.

2. The anion from the salt is needed to partially neutralize added acid.

3. The cation from the salt is needed to partially neutralize added acid.

4. The cation from the salt is needed to partially neutralize added base.

5. Actually, a weak acid by itself is a buffer; no salt is needed. 

Watch Solution

A solution contains 1.00 mole of acetic acid and has pH = 2.50. How many grams of NaCH3COO·3H2O (of MW 136) should be dissolved in the solution to raise the pH to 4.00?

1. 10 g

2. 35 g

3. 20 g

4. 15 g

5. 24 g

Watch Solution

You add a small amount of HCl to a solution of 0.20 M HBrO and 0.20 M NaBrO. What do you expect to happen?

1. The [OH] will decrease slightly.

2. The H+ ions will react with the HBrO molecules.

3. The pH will increase slightly.

4. The Ka for HBrO will increase. 

Watch Solution

You wish to prepare an HC 2H3O2 buffer with a pH of 4.24. If the pK a of is 4.74, what ratio of C2H3O2- / HC2H3O2 must you use?

A) 0.10

B) 0.50

C) 0.32

D) 2.0

E) 2.8

Watch Solution

Identify a good buffer.

A) small amounts of both a weak acid and its conjugate base

B) significant amounts of both a strong acid and a strong base

C) small amounts of both a strong acid and a strong base

D) significant amount of both a weak acid and a strong acid

E) significant amounts of both a weak acid and its conjugate base

Watch Solution

Two compounds, HA & HB have the same K a value. If a solution is prepared containing 0.5  mole of HA and 0.5 of NaB which of the following is True?

A) The solution pH = pKa but it has no Buffer Capacity

B) The solution is not a Buffer because there is no conjugate base of A or conjugate acid of B-.

C) This solution is a good Buffer because the [weak base] = [weak acid]

D) The pH = pKb for the B-

E)  It is not possible to determine if this is a good buffer or not.

Watch Solution

Calculate the pH of a solution after adding 28.522 mL of 0.285 M NaOH to 32.817 mL of a 0.318 M benzoic acid solution. Ka of benzoic acid is 6.5 x 10-5

A. 4.08
B. 4.74
C. 3.64
D. 3.97
E. 4.19

Watch Solution

Consider a buffer composed of a mixture of acetic acid and sodium acetate in equal amounts. Which of the following clearly indicates what happens to the buffer when a very small amount of concentrated LiOH is added to it.


   [CH3COOH]      [CH 3COO – ]          pH
A. increases          decreases        decreases
B. decreases         increases          increases
C. increases          decreases         increases
D. decreases         increases          decreases
E. increases          increases           increases

Watch Solution

Which of the following weak acids would be BEST to use to make a buffer of

pH = 3.52?

a. nitrous acid

b. hydrofluoric acid

c. chloroacetic acid

d. lactic acid

e. benzoic acid

Watch Solution

Which of the following weak acids would be BEST to use to make a buffer of pH = 2.90

a. lactic
b. butanoic
c. chloroacetic
d. benzoic
e. hypochlorous

Watch Solution

Calculate the pH of a solution created by mixing 168.9 mL of 0.395 M Butanoic acid, HC4H7O2, with 293.8 mL of 0.189 M Potassium hydroxide, KOH. The K  a of butanoic acid is 1.5 x 10-5

a. 5.52

b. 5.14

c. 4.90

d. 4.74

e. 4.50

Watch Solution

How much sodium lactate NaC3H3O3, should be added to 583.58 mL of a 0.473 M lactic acid, HC3H3O3, solution to make a buffer of pH 4.08? (assume any change in volume is insignificant and that  the pKa is rounded to two significant figures)

a. 6.986 grams

b. 1.157 grams

c. 127.4 grams

d. 17.89 grams

e. 51.59 grams

Watch Solution

What is the pH of a solution which is 0.400 M in dimethylamine (CH 3)2NH) and 0.600 M in dimethylamine hydrochloride (CH3)2NH2+Cl)? Kb for dimethylamine = 0.00074.

1. 10.78

2. 2.95

3. 10.87

4. 3.31

5. 10.69 

6. 11.05

7. 11.21

Watch Solution

Which one of the following pairs of substances cannot be mixed together (in any quantities) to form a buffer solution?

(a) H3PO4, KH2PO4

(b) NH3, NH4Cl

(c) HNO2, NaNO2

(d) CH3COONa, HCl

(e) HNO3, KOH

Watch Solution

Which one of the following pairs of substances cannot be mixed together (in any quantities) to form a buffer solution?

(a) H3PO4, KH2PO4

(b) NH3, NH4Cl

(c) HNO2, NaNO2

(d) CH3COONa, CH3COOHa

(e) HNO3, KOH

 

Watch Solution

If a biochemist wishes to prepare a buffer that will be effective at a pH of 3.0 (at 25°C) for studying peptide degradation, what will be the best choice for the acid component?

(a) Oxalic acid (H2C2O4), Ka = 5.9 x 10 -2

(b) Hydrofluoric acid (HF), Ka = 6.6 x 10 -4

(c) Pyridinium ion (HC5H5N+ ), Ka = 5.6 x 10 -6

(d) Hypochlorous acid (HClO), Ka = 3.0 x 10 -8

(e) Hydrocyanic acid (HCN), Ka = 6.17 x 10 -10 

Watch Solution

What is the pH of a solution of 0.46 M acid and 0.36 M of its conjugate base if the pKa is 5.51?

a) 4.70

b) 4.90

c) 5.20

d) 5.40

e) 5.62

Watch Solution

A student is asked to prepare a buffer solution at pH = 8.60 using one of the following weak acids: HA (Ka = 2.7 x 10-3), HB(Ka= 4.4x10-6), HC (Ka = 2.6 x 10-9), Which acid should she choose?

Watch Solution

What is the pH of a 0.20 M NH 3/0.20 M NH4Cl buffer after 10 mL of 0.10 M HCl has been added to 65 mL of buffer?

Watch Solution

The pH of a sodium acetate-acetic acid buffer is 4.50. Calculate the ratio of [CH3COO- ]/[CH3COOH].

Watch Solution

Which one of the following pairs, when mixed, do not form a buffered solution?

a) 1.0 M CH3CH2NH2/1.0 M CH3CH2NH3Cl

b) 1.0 M CH3CH2COOH/1.0 M CH3CH2COONa

c) 1.0 M KOH/1.0 M HI

d) 1.0 M NH3/1.0 M NH4Cl

e) 1.0 M H3PO4/1.0 M KH2PO4

Watch Solution

What is the pH of a 500 mL buffer containing 0.10 M Na 2HPO4/0.15 M KH2POafter adding 19 mL of 1 M NaOH?

Watch Solution

Which of the following acids and their conjugate base would form a buffer with a pH of 8.10?

  1. HClO3 pKa = 7.54
  2. H2SO3 pKa = 1.77
  3. HC2H3O2 pKa = 4.74
  4. HNO2 pKa = 3.34
  5. HIO pKa = 10.64
Watch Solution

Calculate the pH of a solution that is 0.210M in nitrous acid and 0.290M in potassium nitrite. The acid dissociation constant of nitrous acid is 4.50x10 –4.

  1. 3.21
  2. 3.49
  3. 13.86
  4. 10.51
  5. 4.56
Watch Solution

Calculate the pH of a buffer that is 0.145M in acetic acid and 0.202M in sodium acetate. The acid dissociation constant for acetic acid 1.75x10 –5.

  1. 4.60
  2. 4.89
  3. 5.05
  4. 4.74
  5. 9.01
Watch Solution

Which of the following acids and their conjugate base would form a buffer with pH of 3.34?

  1. HClO Ka = 2.9 x 10 –8
  2. C6H5CO2H Ka = 6.5 x 10 –5
  3. HF Ka = 3.5 x 10 –4
  4. HIO3 Ka = 1.7 x 10 –1
  5. HClO2 Ka = 1.1 x 10 –2
Watch Solution

The Henderson–Hasselbalch (H-H) equation is a clever rearrangement of the  Ka equilibrium expression that comes in very handy for certain pH calculations. The H-H equation can be used to calculate the buffer ratio ([base]/[acid]) required to prepare a buffer at a specific pH. It is possible that the buffer ratio required to prepare a pH 6.85 buffer could be used to study novel chemotherapy agents; however, the noble H-H equation can be used for more nefarious purposes.

Watch Solution

Which of the following solutions is a good buffer system?

A) A solution that is 0.10 M NaCl and 0.10 M HCl

B) A solution that is 0.10 M HCN and 0.10 M LiCN

C) A solution that is 0.10 M NaOH and 0.10 M HNO 3

D) A solution that is 0.10 M HNO3 and 0.10 M NaNO 3

E) A solution that is 0.10 M HCN and 0.10 M KI

Watch Solution

Which of the following is TRUE?

A) An effective buffer has a [base]/[acid] ratio in the range of 10 - 100.

B) A buffer is most resistant to pH change when [acid] = [conjugate base]

C) An effective buffer has very small absolute concentrations of acid and conjugate base.

D) A buffer can not be destroyed by adding too much strong base. It can only be  destroyed by adding too much strong acid.

E) None of the above are true.

 

Watch Solution

Calculate the pH of a buffer that is 0.225 M HC 2H3Oand 0.162 M KC2H3O2. The Ka for HC2H3O2 is 1.8 × 10-5

A) 4.89

B) 9.11

C) 4.74

D) 9.26

E) 4.60

Watch Solution

How many moles of solid NaF would have to be added to 1.0 L of 1.90   M HF solution to achieve a pH of 3.35? Assume there is no volume change.

(Ka for HF = 7.2 x 10 -4)     

 

Watch Solution

Indicate True or False for the questions below regarding a  0.100 M solution of NH 3

A.  This solution would be a buffer.

B.  This solute acts as a Lewis base.

C.  The pH relevant reaction is: NH 3(aq) + H2O(l) → NH2-(aq) + H3O+(aq). 

D.  Measuring 0.42585 grams of NH 3 and adding water until you reach 250 mL would make this solution. 

E.  There is hydrogen bonding between NH3 and H2O.  

F.  NH3 is trigonal pyramidal.  

Watch Solution

A solution was made by dissolving 0.1219 moles of sodium acetate in 200.0 mL of  1.00 M acetic acid.  Assuming the change in volume when the sodium acetate is not significant, estimate the pH of the solution. The Ka for acetic acid is 1.7 x 10-5

  A.  5.42

  B.  7.31

  C.  9.42

  D.  4.56

  E.  2.89

Watch Solution

How many moles of solid NaF would have to be added to 1.0 L of 1.90 M HF solution to achieve a buffer of pH 3.35? Assume there is no volume change.

(Ka for HF = 7.2 x 10 -4)

A. 3.1

B. 2.3

C. 1.6

D. 1.0

E. 4.9 

Watch Solution

Consider a solution consisting of the following two buffer systems:

H2CO3    ⇌ HCO3 - + H +          pKa = 6.4

H2PO4 -   ⇌ HPO4-2 + H +         pKa = 7.2

At pH 6.4, which one of the following is TRUE of the relative amounts of acid and conjugate base present?

A) [H2CO3] = [HCO3 - ] and [H2PO4 - ] > [HPO4 -2 ]

B) [H2CO3] > [HCO3 - ] and [H2PO4 - ] > [HPO4 -2 ]

C) [H2CO3] = [HCO3 ] and [HPO4 -2 ] > [H2PO4 ]

D) [HCO3 ] > [H2CO3] and [HPO4 -2 ] > [H2PO4

E) [H2CO3] > [HCO3 - ] and [HPO4 -2] > [H2PO4 -]

Watch Solution

What is the pH of a solution that is 0.2 M in acetic acid (Ka= 1.8 x 10  -5) and 0.2 M in sodium acetate?

A. 4.7

B. 9.3

C. 7.0

D. 5.4

E. 8.6

Watch Solution

Which of the following combinations can result in the formation of a buffer?

a) 90 mL of 0.10 M HCN with 70 mL of 0.10 M NH  3.

b) 70 mL of 0.10 M HC 2H3O2 with 30.0 mL of 0.15 M Ba(OH) 2.

c) 50 mL of 0.10 M HNO3 with 50 mL of 0.10 M NaOH.

d) 70 mL of 0.10 M HNO2 with 25 mL of 0.10 M LiH.

e) 70 mL of 0.10 M HClO with 70 mL of 0.10 M LiOH. 

Watch Solution

To create a buffer that maintains a pH around 7.54, which solution would you choose?

A. CH3COOH and NaCH3COO

B. HClO and KClO

C. NaOH and HCN

D. HNO3 and KNO3

Watch Solution

Which of the following combinations can result in the formation of a buffer?

a) HBr and NH4Cl

b) LiOH and HNO3

c) H3PO4 and LiH2PO4

e) None of these combinations result in a buffer.

Watch Solution

Which of the following is TRUE?

a) An effective buffer has a [base]/[acid] ratio in the range of 10 - 100.

b) A buffer is most resistant to pH change when [weak acid] = [conjugate base]

c) An effective buffer has very small absolute concentrations of weak acid and conjugate base.

d) A buffer cannot be destroyed by adding too much strong base, it can only be destroyed by adding too much strong acid. 

e) None of the above are true.

Watch Solution

Which weak acid-conjugate base combination would be the best to create a solution with a pH of 7.538?

a) Butanoic acid and Potassium butanoate (Ka = 1.5 x 10 -5)

b) Benzoic acid and Sodium benzoate (Ka = 6.3 x 10 -5)

c) Acetic acid and Lithium acetate (Ka = 1.8 x 10 -5)

d) Ammonium chloride and Ammonia (Kb = 1.8 x 10 -5)

e) Hypochlorous acid and sodium hypochlorite (Ka = 2.9 x 10 -8)

Watch Solution

A buffer solution is comprised of 50.0 mL of a 0.100 M HC 2H3O2 and 60.0 mL of a 0.100 M NaC2H3O2. Which of the following actions would completely destroy the buffer?

a) Adding 0.003 mol HC2H3O2

b) Adding 0.007 mol Ca(C2H3O2)2

c) Adding 0.005 mol NaH

d) Adding 0.004 mol HNO3

Watch Solution

A buffer:

a. can be made by a combination of a strong acid and strong base.

b. can be a combination of a weak acid and its conjugate base.

c. resists changes in pH.

d. both A and C

e. both B and C

Watch Solution

An aqueous solution contains 0.26M hydrocyanic acid, HCN. One liter of this solution could be converted to a buffer by the addition of:

 

 

Watch Solution

A buffer is prepared using 0.085M formic acid and 0.085M sodium formate. Which of the following statements is false?

 

Watch Solution

Which of the following acids and their conjugate base would form a buffer with a pH of 8.10?

 

A. HClO3 pKa= 7.54

B. H2SO3 pKa= 1.77

C. HC2H3O2 pKa= 4.74

D. HNO2 pKa= 3.34

E. HIO pKa= 10.64

Watch Solution