Properties of buffers
Buffer solutions contain high concentrations of both a weak acid and its conjugate base (or a weak base and its conjugate acid). Because these components can neutralize added H⁺ or OH⁻, buffers are highly resistant to changes in pH. Created by Jay.
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- what if we use strong acid in place of a weak acid(3 votes)
- The conjugate base of a strong acid would be an exceedingly weak base and so it would be severely limited in neutralizing additional acid. Essentially the addition of more acid would lower the pH even more so. The purpose of a buffer is to resist these drastic changes in pH so a strong acid buffer wouldn't really be achieving that.
All you'd have really is just an acidic solution capable of neutralizing additional base only.
Hope that helps.(5 votes)
- How come H+A->HA goes to completion (charges not written)? isn't A- a weak base?1:41(2 votes)
- Well I wouldn’t describe that reaction as going to completion. That would mean the entire supply of conjugate base was converted to weak acid, but we still have leftover conjugate base.
I think what you’re trying to ask is why does the conjugate base neutralize all of the added acid if it’s only a weak acid. A buffer solution is an equilibrium between a weak acid and its conjugate base essentially. What’s happened is that a portion of the conjugate base has neutralized added acid to form new weak acid. These simultaneous changes make sure equilibrium is maintained.
Another way to think about is that the reaction initially exists in equilibrium, and then we add additional acid which disturbs that equilibrium. The reaction moves in a direction to counter that change and restore equilibrium according to Le Chatelier's principal.
Hope that helps.(1 vote)
- So I am struggling with the Henderson-Hasselbalch equation, and it wants me to know how to find the logarithm. It gives an example which is:
H2CO3 ←→ H+ + HCO3-
Ka = [H+][HCO3-]/[H2CO3] = 4.47*10^-7
This can be written as ...
4.47 = 10^.65 because o.65 is the logarithm of 4.47
my question and what has stumped me the most on this is how do you find that the Ka mentioned equals 4.47*10^-7 is there some equation or PH chart to figure that out?
any help would be awesome thanks.(1 vote)
- It's not entirely clear what you're asking. Can you rephrase your question?(1 vote)
- What is the exponential formula used when applying nitrate solutes to the titrations.(1 vote)
- Doesn't the weak acid doesn't lower the pH?(1 vote)
- The weak acid may cause the pH of the buffer solution to be less than 7, but the buffer solution still resists changes in pH.(1 vote)
- Why does the OH ion react with the weak acid but not the H+ ions that would be present in the water due to dissociation? For example, the dissociation of an Alka Seltzer tablet. Wouldn't adding the OH react with the H+?(1 vote)
- I'll assume you mean the H+ (or H3O+, or hydronium) which results from water autoionization.
If we add a base like hydroxide (OH-) to a buffer solution like this, it will react with any acids in the solution. So, both the weak acid and the hydronium from water will react with hydroxide. However, the concentration of hydronium from water is so small compared to that from the acid buffer so it plays no significant role in the pH. We only focus on the react of acid/base with the buffer since it predominantly decides the properties of the solution.
Hope that helps.(1 vote)
- [Instructor] A buffer solution consists of a significant amount of a weak acid and its conjugate base. Let's say we have a generic weak acid, HA, and its conjugate base, A-. And we're gonna use some particulate diagrams to try to understand how buffers work. So for our first particulate diagram, let's count out how many particles we have of each. So first let's count how many HAs we have. So there's 1, 2, 3, 4, 5 HAs, and there's also, for A-, there's 1, 2, 3, 4, 5 A minuses. And when looking at a particulate diagram of a buffer, usually water molecules are omitted for clarity. And also keep in mind that this particulate diagram is just meant to represent a small portion of the solution, so we can get an idea about what's happening in the entire solution. Also notice that like water molecules, cations are also left out of the particulate diagram. So we have five HA particles and five A- particles in our aqueous solution. Having equal amounts of a weak acid and its conjugate base is a good buffer solution. Let's see what happens to the buffer solution if we add in a small amount of acid. So here I'm drawing in an H+ ion, and let's think about adding this H+ ion to our buffer solution. When the H+ ion is added to the solution, the base that is present will react with the H+ ion to neutralize it. So the added H+ reacts with A- to form HA. So for the particulate diagrams, this added H+ is going to react with one of the A minuses present in the buffer solution. So the H+ and the A- form an HA. So we're gonna go from five HAs to six HAs. So let's look at this next particulate diagram here. We can see there are now six HAs in the solution, so let me write down six here. And since we started with five, A minuses and we lost one, we should have only four A minuses in solution now. So let's write down a four here. So you started off with five HAs and five A minuses, and upon the addition of a small amount of acid, the acid was neutralized by the base that was present, and we formed six HAs and four A minuses. So a buffer solution resists changes in pH. So the added H+ was neutralized by the presence of the base. If the buffer solution had not been present, if we just had some water and we added some H+, the pH would have changed dramatically. One way to write the acid-base neutralization reaction that occurred is to write H+ plus A- goes to HA. However, since H3O+ and H+ are used interchangeably in chemistry, we could have also written the net ionic equation as H3O+ plus A- goes to HA and H2O. Next, let's go back to our middle particulate diagram with five HA and five A-. And this time let's try to add some hydroxide ions to the solution. So think about, let me go ahead and draw an arrow here, so we're gonna add a small amount of base to our buffer solution. The hydroxide anion will react with the weak acid that is present, HA, to form H2O and A-. So for the particulate diagrams, we can think about this OH- reacting with one of the HAs to form H2O and an A-. Since we're using up one of the HAs, we're gonna go from five HAs down to four HAs. So let me write down here, four HAs. And when this HA reacts, it's gonna turn into an A-. So we're gonna go from five A minuses up to six. So let's count them over here. 1, 2, 3, 4, 5, 6. So let's write in six A minuses. If there is no buffer present and we're just adding hydroxide anions to water, the pH would change dramatically. However, with a buffer present, since there is a weak acid, HA, present to neutralize the added hydroxide anions, the buffer solution resists a change to the pH. So let's summarize how buffer solutions work. If we add a small amount of an acid, H+, to a buffer solution, the conjugate base that's present, A-, neutralizes the added acid. Therefore, the buffer solution resists a change in pH. And if we add a small amount of a base, the weak acid that's present will neutralize the hydroxide anions. Therefore, the buffer solution resists a change in pH. Let's say we have an aqueous solution of acetic acid and an aqueous solution of sodium acetate. And let's say we have equal moles of acetic acid and of sodium acetate. Mixing these two solutions together would form a buffer solution. To understand why this forms a buffer solution, let's first think about acetic acid. Acetic acid is a weak acid and only partially ionizes in aqueous solution. Therefore, in aqueous solution, we have mostly acetic acid, CH3COOH. Sodium acetate is a soluble salt and it dissociates completely in aqueous solution. Therefore, in aqueous solution, we have sodium cations and acetate anions, CH3COO-. The acetate anion is the conjugate base to acetic acid. Therefore, in aqueous solution, we have a weak acid and its conjugate base, and so we have a buffer solution. If we try adding a small amount of acid to the buffer solution, the conjugate base that's present will react with the acid and neutralize it. So in the balanced net ionic equation, the added acid hydronium ion, H3O+, reacts with the acetate anion to form acetic acid and water. So the added hydronium ion was neutralized, and the buffer has resisted a dramatic change in pH. If we try adding a small amount of base to the buffer solution, the weak acid that is present will neutralize the added base. So in the balanced net ionic equation, hydroxide anions react with acetic acid to form the acetate anion and water. So the added base was neutralized. So to summarize, the acetic acid-acetate buffer system resists dramatic changes in pH when small amounts of acid or base are added. And I just noticed, I forgot to include the L for liquid for water in the balanced net ionic equations.