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## MCAT

### Course: MCAT > Unit 9

Lesson 1: Acid/base equilibria- Acid/base questions
- Acid-base definitions
- Chemistry of buffers and buffers in our blood
- Ka and acid strength
- Autoionization of water
- Definition of pH
- Strong acid solutions
- Strong base solutions
- Weak acid equilibrium
- Weak base equilibrium
- Relationship between Ka and Kb
- Acid–base properties of salts
- pH of salt solutions
- Common ion effect and buffers
- Buffer solutions
- Buffer solution pH calculations

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# Common ion effect and buffers

The common ion effect describes the effect on equilibrium that occurs when a common ion (an ion that is already contained in the solution) is added to a solution. The common ion effect generally decreases solubility of a solute. It also can have an effect on buffering solutions, as adding more conjugate ions may shift the pH of the solution.

## Want to join the conversation?

- At11:35, why is the x in the numerator (outside the bracket) left while the other x's are neglected taking into consideration that all the values of the x are very small and are neglected everywhere else?(14 votes)
- Because you're multiplying the x outside the bracket (see11:10), not adding or subtracting. Multiplying or dividing by 0.35 is considerable compared to adding or subtracting a tiny value from 0.35.(24 votes)

- at3:17, what is the dissociation constant of CH3COONa? How do we know that it completely dissolves and there is 1M CH3COO- and 1M Na+?(14 votes)
- When an ionic bond breaks down, it dissociates completely. When a compound breaks down in an acid/base reaction, it may not be complete because the water is needed to pull off the H+ to form hydronium, but in the case of Na+ it will completely fall off(17 votes)

- ***

In the first example, why doesn't the Na+ get into the equilibrium formula?

Like:

CH3COOH + H2O <---> (CH3COO-) + (H3O+) +*(Na+)*(9 votes)- The sodium is known as a "spectator ion." This means that it does not react with anything but remains present in the solution. The correct way to represent the equation would be:

CH3COOH + H2O + (Na+) <---> (CH3COO-) + (H3O+) + (Na+)(17 votes)

- At 3.56, he accounted for the 1M CH3COOH by subtracting x, but didn't account for the 1M for CH3COO-. Over time, wouldn't the reverse reaction be more favorable and require it be included in the ICE equation?(10 votes)
- After some time, when the reaction approaches equilibrium, the reverse reaction will start having a larger effect until it is equal to the forward reaction, but it will never go beyond this equilibrium (if you don't add any additional chemicals.)

When you continue watching you can see what he does with the formula.*I_, the initial concentrations, are under the fraction, while _E*, the final concentrations, are written above the fraction. The reason why H2O is excluded is because water doesn't have concentration (it's "dissolved" in itself in some way)(3 votes)

- At5:17why is x assumed to be much smaller than 1? How would I know to make this assumption in other problems?(5 votes)
- Chemists are lazy creatures.

They assume that x≪1 in order to avoid having to solve a quadratic equation.

But they always have to check whether the assumption is valid.

A common rule of thumb is that x is negligible if the initial concentration of HA divided by Ka is greater than 400.(10 votes)

- At9:25why is the [NH3} a reactant instead of a product. Is it because we are using Kb instead of Ka?(5 votes)
- Why isn't the reaction between water and NO3- taken into account? Doesn't NH4NOE dissociate, and cause NO3- to raise the concentration of OH- since it acts as a base, and thus change the initial concentration of OH- on the ICE table to 0.35M?(5 votes)
- For the second example problem pertaining NH3 and NH4+NO3-, instead of having the NH3 react with water to form NH4+ and -OH, I had NH4+ react with water to form H3O+ and NH3.

NH4+ + H2O <-----> H3O+ + NH3

Initial concentration = 0.35 nothing 0 0.15

Change in concentration = -x nothing +x +x

End concentration = 0.35-x nothing x 0.15+x

I then calculated Ka using (Ka)(Kb)=(Kw), with the given Kb value. Ka = 5.56x10^-10 = [h3o+][nh3]/[nh4+] = (x)(.15+x)/(.35-x).... and solve for x, which = 1.3x10^-9; then solve for pH using x = 8.88; which is exactly what was obtained in the video.

Question is, will you also obtain the correct pH value regardless which compound you use to react with water as long as you set up the reaction equation properly to either form H3O+ or -OH and use the Ka value for H3O+ and Kb value for -OH in order to solve for x and then further solve for pH or pOH? Or do you have to use the molecule that doesn't form a by-product; CH3COOH vs CH3COONa, Na+ being considered the by-product, therefore CH3COOH favored; likewise, NH3 favored over NH4+ with by-product NO3-?(4 votes)- As long as the amount of correct OH- or H3O+ is calculated, then the result will always be the same.(2 votes)

- How does adding sodium acetate increase the concentration of acetate ion and decrease the hydronium concentration. I understand the equilibrium shifted to the left because of le chatliers rule but what does that have to do with the change in concentrations?(2 votes)
- Sodium acetate is completely soluble in water, right? So if we add acetate to water, the following equation happens: { CH3COOH + H2O <==> CH3COO- + H3O+ }. You know this already, because you said the equilibrium shifts to the left, which is correct. So, if I'm trying to take CH3COO- and turn it into CH3COOH, I need to pull one of the hydrogens off of hydronium, and stick it on the CH3COO- to make CH3COOH. Then I end up with CH3COOH and H2O. This decreases the concentration of hydronium, because we turned a bunch of hydronium into water. Hopefully that helps!(6 votes)

- I don't get how he knows which one he is writing the equation for in #2. Like why didn't he write it for NH4NO3?(4 votes)

## Video transcript

- [Voiceover] Let's say we
have a solution of acetic acid. And we know what's going
to happen in solution. Acetic acid's going to donate a proton to H2O to form H3O+ or hydronium and the conjugate base to
acetic acid, which is acetate. CH3COO-. And so there's a concentration of acetate anions in solution. What would happen now if you
added some sodium acetate? Right, so if you added
some sodium acetate. Let me go ahead and write this out here. So CH3COO- and Na+. So if you add some sodium
acetate to your solution, you now have some more acetate anions. So you're increasing the concentration of one of your products. You're increasing the concentration
of your acetate anion. And according to Le Chatelier's principle, if you increase the concentration
of one of your products, the equilibrium shifts to the left. So the equilibrium is
going to shift to the left, and that means that some
of this acetate anion is going to react with
some of your hydronium ion when your equilibrium shifts to the left. This decreases the
concentration of hydronium ion, and if you decrease the
concentration of hydronium ion, you're going to increase the
pH of your resulting solution. So the acetate anion is the common ion, and this is the common ion effect. So there are two sources
for your acetate anion. So one is the ionization of acetic acid, that's one source for your acetate anion. The other source is the sodium
acetate that you added in. So we have two sources. So we'd expect a pH that's higher than just a solution of acetic acid alone. Let's go ahead and do the calculation and see that that is true. So calculate the pH of a solution that is one molar in acetic acid. And there's the Ka for acetic acid. And one molar for sodium acetate. So we're just gonna start by rewriting our acid-base reaction. So we have acetic acid plus water, so we're going to have
everything at equilibrium, with our products H3O+
and acetate, CH3COO-. And right now let's just
for a second pretend like we have only acetic acid. So these are the kinds of problems that we've been doing. We started with our initial concentration, here we have a 1.00 molar
concentration of acetic acid. And then for our change, we said whatever concentration
we lose for acetic acid, we would gain for the acetate anion, and therefore we would
gain for the hydronium ion. So we're going to gain a concentration of the acetate anion here. But we have to add in something, right? Because we're also dealing with this, one molar concentration of sodium acetate. So we have another source
for acetate anions. So really we have one molar, let's go ahead and put
in that concentration. So the acetate ions come from two sources. Once again, one is the
ionization of acetic acid. And one is the sodium
acetate that you add in to make your solution. That's how I like to think about it. So your initial concentrations would be one molar for acetic acid, one molar for the acetate anion, and then pretty close to
zero here for hydronium. So at equilibrium, let's go ahead and write what we would have. So at equilibrium we'd have one minus x for our concentration of acetic acid, x for our concentration of hydronium, and one plus x for the
concentration of acetate. Once again, two sources for
our concentration of acetate. We can write a Ka expression. So an equilibrium expression using Ka because we're dealing with acetic acid donating a proton to water here. So Ka is equal to concentration
of products over reactants. So H3O+ times concentration of acetate, so times concentration of acetate, all over the concentration
of our reactants, leaving water out. So the concentration of CH3COOH. So for hydronium, at equilibrium
our concentration is x. So we put an x into here. For acetate, our concentration
of acetate at equilibrium would be one plus x. So let's put that in. So we have 1.00 plus x. And then this would all be over the concentration of acetic acid, which would be one minus x. So over here, we put one minus x. Alright, let's get some
more room down here so we can talk about this. We're going to make that same assumption that we've done before, that this concentration x is
much, much smaller than one. And if that's a really small number, one plus x is pretty much the same as one. And one minus x is also
pretty much the same as one. So if this is a very small concentration, we don't have to worry about these. We can approximate them and say one plus x is equal to one. Let me go ahead and rewrite this here. This would be x times, over here one plus x is
approximately equal to one. So that's our approximation. Over here, one minus
x, if x is very small, is also approximately equal to one. This is all equal to the Ka value, which for acetic acid is 1.8 times 10 to the negative five. And so the ones obviously would
cancel each other out here. So the concentration of
hydronium, which is x, remember x represents the
concentration of hydronium ions, is 1.8 times 10 to the negative five. So to find the pH, all we have to do is take the negative log of that. So negative log of 1.8 times
10 to the negative five. So we can go ahead and get out
the calculator and do that. Negative log of 1.8 times
10 to the negative five is equal to 4.74. So our pH is equal to 4.74. Now if you go back to the
video on weak acid equilibrium, we calculated the pH for
a 1.00 molar solution of acetic acid only, and the pH for that came out to be 2.38. This is because we didn't have any other acetate anions present here. So we have different pHs right? The pH is higher for this situation, where we have acetic acid, where we have two sources
again for the acetate ion, the ionization of acetic acid and also the addition of sodium acetate. Let's do another common ion problem here. So this time we need to calculate the pH of a solution that is
0.15 molar for ammonia and 0.35 molar for ammonium nitrate. So ammonium nitrate is
NH4+ and NO3- in solution. And so we start with ammonia. So let's go ahead and write
what's going to happen when ammonia reacts with water. So ammonia is a weak base, and it's going to take
a proton from water. So if NH3 picks up an H+, we form NH4+ or ammonium. And if we take a proton away from H2O, take an H+ away from H2O, we form OH- or the hydroxide ion. So once again we start with
our initial concentration. And we're going to pretend
like it's one of the problems that we've been doing in earlier videos. So if we have 0.15 molar
concentration of ammonia, we go ahead and put 0.15 here. We think about the change. Whatever concentration we lose for ammonia is the same concentration
that we gain for ammonium since ammonia turns into ammonium. And therefore that's also
the same concentration we gain for hydroxide. So one source for the ammonium ion it would be the protonation of ammonia. So that's one source. But we have an additional source because we also have 0.35
molar ammonium nitrate. So there's another
source for ammonium ions. So we have 0.35 molar, so we go ahead and put 0.35 molar in here for the initial concentration of ammonium. And we're saying that we have zero for our initial
concentration of hydroxide. So when the reaction
comes to equilibrium here, for ammonia we would have 0.15 minus x. For ammonium, we have two sources right? So this is a common ion here. So we have 0.35 plus x. And then for hydroxide
we would have just x. So since ammonia is acting
as a weak base here, let's go ahead and write
our equilibrium expression. And we would write Kb. And Kb for ammonia is 1.8
times 10 to the negative five. So this is equal to concentration of our
products over reactants. So we have concentration of NH4+ times the concentration of OH- all over the concentration
of our reactants, leaving out water. So we just have ammonia here, so concentration of NH3. So let's go ahead and
plug in what we have. For the concentration of ammonium, we have 0.35 plus x. So we put 0.35 plus x. For the concentration of hydroxide, we have x. So we go ahead and put an x in here. And then that's all over the concentration of ammonia at equilibrium, and we go over here, and for ammonia at
equilibrium it's 0.15 minus x. So we write over here 0.15 minus x. And we can plug in the Kb value, 1.8 times 10 to the negative five. So let's go ahead and plug in the Kb. So we have 1.8 times
10 to the negative five is equal to, okay once again, we're going to make the assumption. So if we say that x is
extremely small number, then we don't have to worry about it when we're adding it to 0.35. And so we just say this is equal to 0.35. So 0.35 plus x is pretty close to 0.35. So this is times x. And make sure you
understand this x is this x. And then once again, 0.15 minus x, if x is a very small number, that's approximately equal to 0.15. So now, we would have this. And we need to solve for x. So let's go ahead and do that. Let's get out the calculator here. We need to solve for x. So 1.8 times 10 to the negative five times 0.15, and then we need to divide that by 0.35. And that gives us what x is equal to. And so x is equal to 7.7
times 10 to the negative six. So x is equal to 7.7 times
10 to the negative six. x represents the concentration of hydroxide ions in solution. So this would be the
concentration of hydroxide, so 7.7 times 10 to the negative six molar is our concentration of hydroxide. Our problem wanted us to calculate the pH. So if we know the concentration
of hydroxide ions, we can find the pOH by
taking the negative log of the concentration of hydroxide. So negative log of the
concentration of hydroxide ions will give us the pOH. So let's do that. So we have the negative log of 7.7 times 10 to the negative six. And this gives us the pOH,
where we round to 5.11. So the pOH is 5.11. Let's get a little bit more room here. So pOH is equal to 5.11. And then we're home free because pH plus pOH is equal to 14. So we just plug in our
pOH, solve for the pH. pH is equal to 14 minus 5.11, which of course is equal to 8.89. And so we've calculated the
final pH of our solution.