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## Chemistry library

### Course: Chemistry library > Unit 13

Lesson 2: Acid-base equilibria# Weak acid equilibrium

Quick overview of Ka and pKa. Example of calculating the pH of a weak acid solution.

## Want to join the conversation?

- If this is a weak acid, then why does the pH of the solution come out to be 2.38? Wouldn't that then classify it as a strong acid if the pH is that low? I noticed the same thing in the next video as well with the weak base solution having a pH of 11.48. Why are they not closer to the neutral 7 rather than falling toward the ends of the pH scale?(30 votes)
- As if acids and bases weren't tricky enough... When we refer to the
*strength*of an acid/base in chemistry, it has nothing to do with its pH. Acetic acid (vinegar) is relatively harmless in most residential applications (~5%). It's classified as a weak acid. But if you get concentrated acetic acid (>90%), it is highly corrosive and flammable. But even concentrated acetic acid is still classified as weak.

Or take orange juice. A very low pH (pH 1 or 2). You can drink it and it doesn't rot out your stomach. Citric acid is a weak acid. But get the concentration high enough, and it can be dangerous.

Now take a strong acid, hydrochloric acid, HCl. I've spilled HCl on my bare hands before and my hands didn't melt off. I casually rinsed them off in a sink and continued my work. Because the HCl was 0.01 molar (very weak). I'd be singing a different story if it was concentrated HCl.

Now the lesson in all of this: in chemistry, the*strength*of an acid is based on how much of it can be found in the intact acid form (i.e., HX) and how much can be found in the ionized form (i.e., H+ and X-). Strong acids completely* dissociate into H+ and X-.

And to do well in chemistry, you have to memorize these 7 strong acids. They're the only strong acids, everything else is considered a weak acid:

HCl, HBr, HI, HNO3, H2SO4, H3PO4, and HClO4

(correction, H3PO4 is not a strong acid - but HClO3 is!)

*I use the word completely, but in reality, there may be 1 HX for every 10^30 H+ and X-. It's close enough to say they are completely ionized.(90 votes)

- how would we find the square root of 1.8x10^-5 without a calculator(17 votes)
- A trick I personally like to use:

Square roots like even number exponents, so I'm going to rewrite 1.8 * 10^-5 as 18 * 10^-6.

Since 18 is very close to 16, I'm guessing its square root is around 4.1.

Since 10^-6 has a square root of 10^-3, I'm going to multiply that in too.

Answer: Square root of 18 is approximately 4.1 * 10 ^ -3 (real answer is 4.24*10^-3)

That is more than enough accuracy for the MCAT ;)(55 votes)

- in the reaction at3:20, why is the proton donated from the oxygen, forming CH3COO rather than from the CH3, forming CH2COOH? Thanks!(12 votes)
- CH3COO- would be the conjugate base of acetic acid (CH3COOH) and the reason it donates the proton on the oxygen, as opposed to the carbon, is because of electronegativity. The methyl group (CH3) is significantly less electronegative than the oxygen. This is because the carbon has it's octet filled while the electronegative oxygen is looking for more electrons, the easiest way to do this is to donate the proton. The only time you'll see a carbon donate a proton is when there is a carbocation/anion intermediate in mechanisms and reactions.(12 votes)

- at1:20how do you calculate pka without a calculator?(3 votes)
- pKa implies -log (Ka). If your Ka is in the form (m x 10^n), you can pretty accurately estimate the log without a calculator by putting it into the following equation:

n - 0.m(16 votes)

- If pka for methanol=15.54, then pkb=-1.54. Is that possible?(6 votes)
- Yes. pKb=-log Kb. A negative pKb just means that the Kb is larger than 1

( [conjugate acid][OH-]/[base] > 1 )(4 votes)

- When solving the equation at the end, we assume that 1.00-x is 1.00 because x is negligible because it is so small. Why, then, is the numerator not assumed to be zero as well?(5 votes)
- Because (tiny number)/(1- tiny number) is very close to (tiny number)/1, but

(tiny number)/(1-tiny number) is not zero, it is (tiny number)(2 votes)

- Is the pKa of a particular acid only temperature-dependent whereas the pH of its aqueous solution concentration-dependent as well?(3 votes)
- Precisely. The pKa is a constant for a given compound and varies with temperature. pH depends on the pKa of the acid/base but will also depend on concentration of the acids/bases dissolved.(4 votes)

- Very nit picky question, particularly regarding sig figs. I got pH = 2.37 instead of 2.38

Calculations:

pH = -log( (1.8 * 10^-5)^(1/2)) = -(1/2) * log(1.8*10^-5) (by using log exponent rule)

= -(1/2) * (log(1.8) + log(10^-5)) (by using log multiplication/addition rule)

= -(1/2) * (log(1.8) - 5) (log exponent rule)

= 2.37236374744...

= 2.37 (round to 2-sig figs)

the difference between the author's and mine comes because of intermediate rounding.

What's the proper rule for this?(1 vote)- You did it correctly! You didn’t do any intermediate rounding.

The video rounded log[√(1.8×10⁻⁵)] and then rounded this again when calculating the pH.

This is called**accumulated roundoff error**.

You don’t have to write down 12 significant figures for every intermediate answer, but you should carry one extra digit (called a**guard digit**) than is required.

That way, you can avoid accumulated roundoff error.(5 votes)

- What is contrast between Kb and pKb? Can any one explain it to me? Thanks(2 votes)
- The are related as logarithm and antilogarithm.

pKb = -log₁₀(Kb)

log₁₀Kb = -pKb

Kb = antilog[log₁₀(Kb)] = antilog(-pKb) = 10^(-pKb)(3 votes)

- at7:25--

"... we're going to assume that x is << 1."

It does make things easier, but how do we know this assumption is valid?(2 votes)- Solve it as a quadratic afterwards to check and see how valid the result is. If the Ka or Kb is around 10^-4 or smaller range the assumption is generally okay.(2 votes)

## Video transcript

- [Voiceover] We've already talked about how to write an equilibrium expression. So if we have some generic acid HA that donates a proton to H2O, H2O becomes H3O+ and HA turns into the conjugate base which is A minus. And so here's our equilibrium expression and the ionization constant
Ka for a weak acid. We already talked about the fact that it's going to be less than one. So here we have three weak acids. Hydrofluoric acid,
acetic acid and methanol. And over here are the Ka values. So, you can see that a hydrofluoric acid has the largest Ka value. Even though they're all
considered to be weak acids, all right, 3.5 times
10 to the negative four is larger than 1.8 times
10 to the negative five. So hydrofluoric acid is
stronger than acetic acid and acetic acid is stronger than methanol. But again, they're all
considered to be weak acids relative to the stronger ones. So let's talk about pKa. The pKa is defined as the negative log of the Ka. So if we wanted to find
the pKa for methanol all we have to do is take the Ka and take the negative log of it, so that pKa is equal to the negative log. Negative log of 2.9 times 10 to the negative 16. So let's get out the
calculator and let's do that. Negative log of 2.9 times 10 to the negative 16. And this gives us 15.54
when we round that. So the pKa of methanol, the pKa of methanol is equal to 15.54. We could write in a pKa column right here and for methanol it's 15.54. If you did the same
calculation for acetic acid you would get 4.74. And once again, if you did
this for hydrofluoric acid, you would get 3.46. As we go up on our table here, we're increasing in acid strength. So out of our three weak acids, hydrofluoric acid is the strongest so it has the largest value for a Ka but notice it has the
smallest value for the pKa. So the lower the value for pKa, the more acidic your acid. So 3.46 is lower than 4.74 and so hydrofluoric acid is
more acidic than acetic acid. The problem asks us to calculate the pH of a 1 molar solution of vinegar which is acetic acid in water. Let's start by writing
our acid base reaction. We have acetic acid. So CH3COOH plus water and if acetic acid
donates a proton to water, water would turn into
H3O+, the hydronium ion and we would also get the acetate anion. CH3COO minus the conjugate
base to acetic acid. So before we get to this problem, let's just pretend that
we're starting with 100 molecules of acetic acid. So 100 molecules of acetic acid here and let's say none of them have reacted. This is before any reaction has occurred which means we don't
have any of our products. We don't have any hydronium ions, we don't have any acetate anions. And acetic acid is a weak acid. If a strong acid would ionize 100%, so all 100 of those molecules pretty much would ionize. But since acetic acid is weak, let's pretend like only one of those acetic acid molecules
donates a proton to water. So we're gonna lose one
molecule of acetic acid and that's gonna turn
into the acetate anion. So if we lose one molecule of acetic acid, we gain, all right, we
gain one acetate anion. And therefore, if we protonate
one molecule of water with that proton, we're going to gain one hydronium ion. So, we would say that we would have 99 molecules of acetic acid and then we have one hydronium ion and we would have one acetate anion. So that's the kind of thinking that we're going to
apply to concentration. Next, we're gonna start with an initial concentration here. We're starting with a one
molar solution of acetic acid so I can write here 1.00 as
my initial concentration. And just like before,
we're going to pretend like nothing has happened yet. So the concentration of hydronium is zero and the concentration of the
acetate anion is also zero. So next we're gonna
write a C here for change and let's define X as the concentration of acetic acid that reacts. So, if we have a concentration
of acetic acid as being X, let me go ahead and
I'll just write X here, that means we're gong to lose a certain concentration of acetic acid. And therefore, going to gain the same concentration
of the acetate anion and we're going to therefore, gain the same concentration
of the hydronium ion. So, when everything comes to equilibrium, so our equilibrium concentrations would be one minus X for the concentration of acetic acid and for the concentration of hydronium we would have X. And for the concentration
of acetate anion, we would also have X. Now, let's write our
equilibrium expressions. Let's get a little more room down here. Our equilibrium expressions. So Ka would be equal to
concentration of hydronium ion. The concentration of hydronium ion times the concentration of acetate anion. And this is all over, all over the concentration of acetic acid and we leave water out. So all over the concentration
of acetic acid here. All right, let's plug in what we know. So the equilibrium
concentration of hydronium, that's X, right? So that's X. Let's go ahead and put that in here so we can put X in for
concentration of hydronium. And concentration of the acetate anion, that's also X. That's the concentration
of the acetate anion so I put in an X here too. All over the concentration of
acetic acid at equilibrium. That's one minus X. So I put in here 1.00 minus X and this is equal to
the Ka for acetic acid. So we have that on our table above. Let's go back up here to the table that we talked about earlier and here is the Ka for acetic acid. 1.8 times 10 to the negative five. So we plug that in, we plug that in here. Let's plug in 1.8 times
10 to the negative five. Now, at this point you could solve for X but you need to use the quadratic formula. So let's make an assumption at this point and we're going to assume that X, our concentration is much, much smaller than one molar because that makes our
life easier for the math. Because if X is much,
much smaller than one, one minus X is approximately
the same thing as one. So this is an extremely small number, subtracting it from one
isn't gonna do anything. Isn't gonna do much of anything. We're gonna make this assumption and say that one minus X is equal to one and that just makes our lives easier. So now we have 1.8 times
10 to the negative fifth is equal to X squared over one. So all we have to do now is solve for X. Let's get out the
calculator here, turn it on. We just need to take the square root of 1.8 times 10 to the negative five. And so we get .0042. Let's go ahead and write that down here. Let's get some more room. And X is equal to 0.0042. And remember what X represents. X represents the concentration
of hydronium ion. X here represents the concentration of hydronium ion at equilibrium. So this is equal to our
concentration of hydronium ion and now we can calculate the ph which was the whole point of the problem. Let's calculate the pH of our solution. So the pH will be equal
to the negative log of the concentration of hydronium. So the negative log of .0042 and we can use our calculator for that. Let's go ahead and take
the negative log of .0042 and that's going to give us a pH of 2.38. Let me go ahead and write that. So our pH is 2.38. That is the pH of our vinegar solution.