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### Course: AP®︎/College Chemistry>Unit 13

Lesson 1: Buffer solutions

# Buffer solution pH calculations

Example of calculating the pH of a buffer solution using the Henderson-Hasselbalch equation, including the pH of the buffer solution after adding some NaOH.

## Want to join the conversation?

• At 5.38--> NH4+ reacts with OH- to form more NH3. Since it is an equilibrium reaction, why wont it then move backwards to decrease conc of NH3 and increase conc of NH4+?
• The additional OH- is caused by the addition of the strong base. So these additional OH- molecules are the "shock" to the system. The system counteracts this shock by moving to the right of the equation, thus returning the system to back to equilibrium.

So when the reaction moves to the right, equilibrium is restored. The system is then at equilibrium and there is no cause for the system to move "backwards".

Think of it as a single shock and a single response by the system. Otherwise the logic gets a bit circular and by that logic the system is never able to settle.
• This may seem trivial, but at , why is the hydroxide ion written with the charge on the left-hand side, instead of the right? I've seen it that way consistently in these chemistry videos, but never anywhere else.
• It is preferable to put the charge on the atom that has the charge, so we should write ⁻OH or HO⁻.
However, many people still write the formula as OH⁻. It's OK, as long as you remember that the O atom has the charge.
• At NH4Cl is called an acid, but isn't it a salt?
• It is a salt, but NH4+ is ammonium, which is the conjugate acid of ammonia (NH3).
• How would I be able to calculate the pH of a buffer that includes a polyprotic acid and its conjugate base?
For example, if I were to add a certain amount of the polyprotic acid H2A and a certain amount of its conj base A^2- (obviously in the form of a salt, say Na2A), how would I find the pH? Assume I know the ka1 and ka2 values. Now what? Do I use good ole Henderson Hasselbach? Which pKa do I use? Is there another equation to use in these instances?
• You can still use the Henderson Hasselbach equation for a polyprotic (can give more than two hydrogens, hence needs to have two pKa) but might need to do this twice for depending on the concentration of your different constituents. It is a bit more tedious, but otherwise works the same way.
Btw, amino acids (proteins) are polyprotic and depending which solution they are in, they too will either be acidic or basic buffers.
• how can i identify that solution is buffer solution ? And at how does he know that whole of the NH4Cl is going to dissociate into 0.20M of NH4+
• You need to identify the conjugate acids and bases, and I presume that comes with practice. The same way you know that HCl dissolves to form H+ and Cl-, or H2SO4 form 2H+ and (SO4)2-. In this example with NH4Cl, the conjugate acids and bases are NH4+ and Cl-. Notice how also the way the formula is written will help you identify the conjugate acids and bases (acids come first on the left, bases on the right)
• Very basic question here, but what would be a good way to calculate the logarithm without the use of a calculator?
• There are some tricks for special cases, but in the days before everyone had a calculator, students would have looked up the value of a logarithm in a "log book" (a book the lists a bunch of logarithm values). There isn't a good, simple way to accurately calculate logarithms by hand.
• what happens if you add more acid than base and whipe out all the base
• This question deals with the concepts of buffer capacity and buffer range. A buffer will only be able to soak up so much before being overwhelmed. This is known as its capacity. It's the reason why, in order to get the best buffer possible, you want to have roughly equal amounts of the weak acid [HA] and it's conjugate base [A-]. If you have roughly equal amounts of both and relatively large amounts of both, your buffer can handle a lot of extra acid [H+] or base [A-] being added to it before being overwhelmed.

However, you have to be aware of your buffer's pH. If you took a reaction of a weak acid that has a small Ka value, it will only produce some conjugate base and it's pH might be very low, like a 2. You have to add a lot of conjugate base in order to make this be a buffer (remember, we need roughly equal amounts of the acid and its conjugate base, but in this example, the Ka is low, so we a lot of the weak acid and not much of the conjugate base). So, now that we're adding the conjugate base to make sure we have roughly equal amounts, our pH is no longer 2. It might be somewhere like a 5. So now we have a strong buffer with a lot of capacity, but our pH of this buffer solution is very different from the pH of just the weak acid/conjugate base we started with.
• At the end of the video where you are going to find the pH, you plug in values for the NH3 and NH4+, but then you use the values for pKa and pH. Wouldn't you want to use the pKb to find the pOH and then use that value to find the pH? Thank you.
• I think he specifically wrote the equation with NH4+ on the left side because flipping it this way makes it an acid related question with a weak acid (NH4+) and its conjugate base (NH3). Hasselbach's equation works from the perspective of an acid (note that you can see this if you look at the second part of the equation, where you are calculating log[A-][H+]/[HA].
(1 vote)
• I did the exercise without using the Henderson-Hasselbach equation, like it was showed in the last videos. But my thought was like this: the NH4+ would be a conjugate acid, because I was assuming NH3 is a base. So, I would find the concentration of OH- (considering NH3 in an aqueous solution <---> NH4+ + OH- would be formed) and by this, the value of pOH, that should be subtracted by 14 (as pH + pOH = 14).
What did I think wrong? I really got confused... :/
• The 0 isn't the final concentration of OH⁻. The 0 just shows that the OH⁻ provided by NaOH was all used up. In order to find the final concentration, you would need to write down the equilibrium reaction and calculate the final concentrations through Kb.
• Commercial"concentrated hydrochloric acid"is a37%(w/w)solution of HCl in water.(density of HCl is1.017g/mol)calculate the amount of water needed to be added in order to prepare 6.00M of HCl from 2dm3 of the concentrated HCl.
• We know that 37% w/w means that 37g of HCl dissolved in water to make the solution so now using mass and density we will calculate the volume of it. Then by using dilution formula we will calculate the answer.