﻿In this video we'll look at a weak acid equilibrium question; this is an example of an
AP chemistry free response question. Now this one is a little bit longer because I wanted
to get some extra stuff in here, but it's still excellent preparation. So just as a reminder you
can get all sorts of resources for AP chemistry here at my website. OK, so let's get started.
Formic acid reacts with water according to the following equation. This is formic acid
and the equation gives us the reaction for the ionization of formic acid in water, and we get the
Ka, or acid dissociation constant, of formic acid. OK, now we know that formic acid is a weak
acid because it doesn't fully ionize. Instead, the products and the reactants are in equilibrium with
each other. OK now, there are three parts to this question; we'll go through them one at a time,
but before we start, remember that different parts of the AP free response questions sometimes depend
on each other. OK, so, we'll start here with part A. Write the equilibrium constant expression Ka
for formic acid. Equilibrium is a super common topic on the free response section
so many questions begin by asking you to write the equilibrium expression
k. For weak acid, like formic acid here, the k is the acid dissociation constant expression or Ka.
OK, so for any weak acid that dissociates according to this equation the equilibrium
expression will be based on this setup. Let's take a look at the parts here. OK, now remember
that H2O is a liquid, so it doesn't appear in the equilibrium expression. And the AP exam actually
gives you this Ka formula in this sort of general setup on the equations page. And it also gives you
the Kb expression for base questions too. OK, now there's a discrepancy to point out here.
Note that in this equation and in this expression, we're just using H+, but the equation up here has
H3O+, which is hydronium. You should remember that these are just different ways of expressing
the same thing, OK? Hydronium ion is essentially represented here by H+. You'll see both of these
on the AP exam, and you'll see them moving back and forth between hydronium and H+,
so you don't want to get confused. These are essentially interchangeable. OK so here is
our Ka expression for formic acid OK. Our Ka equals the concentration of hydronium, or H+,
times the concentration of the formate ion, OK? The formate ion here is A- right here, so that's
formate, A-. And then finally this is divided by the concentration of formic acid, which is our HA,
OK? So this is our acid dissociation constant expression and if you know a little bit about
the AP exam or even if you just have done weak acid equilibrium problems like this in the past,
you know that we're probably going to be using this to do some more calculations down the road.
OK, we're going to keep it here for just that reason and we'll move on to part B, OK.
Calculate the pH of a 0.12 molar aqueous solution of formic acid. OK, now the pH
that we have to calculate is the negative log of the hydrogen ion, or the hydronium ion
concentration. If you forgot this equation it's right there on the equation sheet,
OK? So to get the pH we have to figure out the H+ concentration here. And just to make things easier
we'll just call it H+ or hydrogen from that one. Now, remember formic acid is a weak acid, OK? So
if we start with a concentration of 0.15 molar, only some of that is going to ionize to make H+,
OK? Not all of it, just some of it. So, to find the concentration for these chemical species at
equilibrium we're going to use an ICE table. This includes all of the relevant participants
in our equilibrium expression and it's going to tell us how many of each we have when this
reaction reaches equilibrium. OK so ICE stands for initial change and equilibrium. The AP exam loves
ICE tables, so make sure you understand the manipulations that we're going to be doing here.
OK, so when we set up a table like this, we can then start putting in the different
entries. Here we'll start with the initial concentrations, OK? First
the initial concentration of formic acid, well that's given to us in the problem, 0.15 molar.
Now we assume that before we reach equilibrium initially H+ and formate are going to be zero
because we're starting just with this, the formic acid on the left side, OK? Now as this
reaction reaches equilibrium, what are the changes that are going to take place here?
OK, well the amount of formic acid is going to decrease by x. It's going to decrease by some
amount because it is turning into these products. OK, so we'll do minus x here, that's the amount
that the formate is going to decrease, and then because the H+ and the formate started at zero,
these concentrations had to increase to reach equilibrium. So if we look at the
stoichiometry of the reaction we can see that every one molecule of formic acid dissociates
to form one H+ and one formate. Alright, that's a one to one to one stoichiometric ratio.
OK, so if this goes down by x, this is going to go up by x, and this is also going to go up by x,
right? 1 to 1 to 1. OK, so this is the change that we'll see and finally here is what we're
going to have at equilibrium. We're going to have the initial concentration of formic acid minus x,
and then we're going to essentially have 0 plus x for H+, and we're going to have 0 plus x
for formate. So this gives us the equilibrium concentrations. We're going to be using this
to solve for H+. We can put these equilibrium values into that Ka expression that we came up
with earlier. OK, so this you can kind of see how part two of the question depends a little bit on
part one. OK, now let's plug in our values. OK, at equilibrium we have H+, which is x,
we have formate, which is also x—so x times x—and on the bottom we have the concentration of formic
acid that we're going to have at equilibrium. Now just to keep in mind why are we doing all this,
we're doing this because we ultimately want to know x, the concentration of H+ at equilibrium.
Then we'll be able to figure out what our pH is. So we're doing all of this to solve for the value
of x to get H+ to then be able to calculate pH. Things are getting a little cluttered here
so let's get rid of some of the stuff that we don't need. Move these equations around
and there we go. Now we need to solve for x. Like many acid-base equilibrium problems
solving for x can seem pretty complicated at first. To solve an equation like this we could
use the quadratic equation, but that takes a lot of time. Instead, we can take advantage of
a shortcut, and we could take advantage of the fact that the Ka value is a very small number.
And that means that x is going to be quite small. It's particularly going to be small
when compared to the concentration of 0.15 molar here. This is going to be way bigger than x, so
for our purposes here we can assume that for this part of the equation the value of x is negligible,
and we can just drop it, OK? And if we do that we get this, OK, the x just drops out here. Now
Ka approximately equals x squared divided by 0.15. This simplifies the math and it makes the
problem a lot easier. And you've probably solved many equilibrium problems like this in class.
The AP exam pretty much just expects that you're going to make this assumption as well. And in
fact, almost all acid base equilibrium problems make this except make this assumption. OK, now,
just remember to write down any assumptions that you make when you're showing your work,
so the test grader understands your reasoning and doesn't think that you just forgot this x here.
Alright, so we want to justify why we're getting rid of it and ending up with this approximation.
So, we can now rearrange this to solve for the equation, to solve the equation.
Alright? So, we will multiply both sides by 0.15 and we'll get x squared equals Ka
times that. Now we want to get rid of x squared, so we'll take the square root of both sides.
That gives us x equals the square root of this expression here. And now we have the Ka.
Well that's pretty easy. We can just plug this value in for the Ka. We get that times that.
Put this whole thing into a calculator ,round to two significant figures, and that gives us 0.0052
molar. OK, that is the value of x. Now remember what we were solving for.
The value of x is the concentration of H+, or the concentration of hydronium. So now
we can take this and we can put it into our pH equation there. We put this in for H+,
and when we do this math, the negative log of the H+ concentration there is our pH value, 2.3.
OK now here is the third part. To create a buffer solution sodium formate can be added to formic
acid. How many moles of sodium formate would need to be added to 250 milliliters of a 0.10 molar
formic acid solution to create a buffer with a ph of 3.90? OK, buffer, this is a buffer problem.
We're going to be adding sodium formate to formic acid to create a buffer with a pH of 3.90.
Now like many buffer problems on the AP chem exam, we're going to solve it
by using the Henderson-Hasselbalch equation. And that's one of the equations that's given to you
on the formula sheet. And here's what it tells you, OK? It says the pH of a buffered solution
is equal to the pKa of the acid plus the log of the conjugate base divided by, or I
should say the log of the concentration of the conjugate base, divided by the concentration of
the acid. And remember for this problem A- is the formate ion, H+ is formic acid. So
we can swap those in right now formate and formic acid. Now what do we know here? What are we
going to be solving for? Alright, we know the pH and we can find the pKa from the Ka.
And we also know the concentration of formic acid, alright. That’s right here,
so the only unknown then is the concentration of the formate ion. Essentially all of this formate
ion is going to be coming from the sodium formate that we're adding here. So once we find this
we can then figure out how many moles of sodium formate we're going to need to add to achieve this
concentration. OK, there's a couple ways that we could rearrange this equation to solve for
formate. And they're all equally valid. You can rearrange this equation however you like. We're
going to take it slow. We're going to step by step here. And the first thing that we're going to do
is we're going to subtract pKa from both sides and then we'll just like flip the equation here
to make it a little easier. So we get log formate divided by formic acid equals pH minus pKa.
OK, and right away we can plug in a couple knowns. We know that the concentration of formic acid
is this and we know that the pH of the buffer is 3.90, so we can go ahead, and we can plug those
values in. There they are now. What about the pKa here? We have to calculate that,
but we can calculate it pretty easily from the Ka because pKa equals the negative log of the Ka.
So, there's our Ka. When we do this math, we get 3.75. Stick it up there and now this is relatively
simple subtraction, 0.15. OK, we can clean this up just a little bit. Let's get rid of that, get rid
of that, and then really the only thing that we're worried about right now is the fact that this
equals 0.15 there we go. Now I'll just move it up, OK. Now remember we're solving for formate here.
We want to get rid of this log function, now how are we going to do this? Well, when we say log, we
really mean log base 10. So, to get rid of the log function we take both sides of the equation and
we place them as exponents on 10. Alright, when we do that, it removes a log function from this side,
10 raised to the log of something is just the something. OK and then over here we have 10
raised to this side of the equation. So that's how we undid the log function,
alright? And now it's going to be relatively simple math to solve. For formate, OK, we can just
multiply this value by this value. There it is. Solve for formate, do some math, 0.14 molar.
OK so our concentration of formate is 0.14 molar, but now we have to finish the problem. How many
moles of sodium formate must be added to a 250 milliliter solution to achieve this concentration?
Let's clear the slate again and list our knowns. OK, the concentration of formate is this, we just
solved for, and our volume is 250 milliliters. So we're going to use the molarity equation
to finish this up molarity equals moles divided by liters of solution. Remember it requires
liters of solution and the problem gives us milliliters so you have to convert. There's
a thousand milliliters in one liter, so you either divide this by a thousand or you just
move the decimal place over three spots to the left in your head. So 250 milliliters equals 0.25
liters, liters down there. We can rearrange that equation, moles equals molarity times
liters and we can solve for moles plugging in these values. Molarity up here, liters up here,
and this is the number of moles of sodium formate that must be added to create this
buffer solution with a final pH of 3.90, OK? Now before we move on notice how part C here
did not depend on parts A and B on the earlier part, so even if you mess something up at earlier
stages as you're working through one of these multi-step problems, don't worry about it,
just keep moving on. Remember to show your work and detail your thinking.
So here we've seen how to tackle some of the most common acid-base equilibrium
problems the AP test is going to ask you. If you want to get a good score
on the test, try to work through as many example problems as you can, and you'll begin to see
that there are patterns and there's specific types of questions that they tend to ask. And
being familiar with these is really going to help you be comfortable on test day. Good luck!