AP Chemistry 2024 – 7.7 Calculating Equilibrium Concentrations | Fiveable (2024)

One of the most important aspects of is that it is the point at which a “settles” and concentrations will not change because the rate of the forward and reverse will equal each other. However, what happens if we know the constant for a , but don’t know what the concentrations for a are? Well, we have to do some math!

In this section, you’ll learn how to use an (sometimes called a ) to solve for concentrations and learn some of the math behind one of AP Chemistry’s most important mathematical techniques.

ICE Boxes Explained

When solving for an concentration, we use an . The “I” stands for “initial”, the “C” stands for “change”, and the “E” stands for “”. If you ever see , the "R" stands for "." Both techniques are the same!

These boxes are used to show the initial concentrations, , and for a . Let’s take a look:

Image From Alibaba

Okay…just kidding. Here’s what a real looks like with the CH₃COOH ⇌ CH₃COO⁻ + H⁺ (K = 1.8 * 10⁻⁵):

Filling in an ICE Table

Let’s start filling this in! Let’s suppose we are starting with an initial CH₃COOH concentration of 1 M (initial concentrations are typically given!). Because we’re starting at the very beginning before any has occurred, the concentration of the products, [CH₃COO⁻] and [H⁺], are 0:

Next, let’s think about how these quantities change. When this occurs, some of our reactants will start converting into products. However, because our K is low, not all of the reactant is going to turn into products, so we don’t know how much we form. Therefore in the change row, we use x as a variable to denote this change. For reactants, we lose nx (where n is the ), and for each product, we gain nx. At , we add together the initial and change to get our final concentrations! Let’s see this in the table:

Note that K is low and we are displaying change, we subtract x for the reactants to note a loss of concentration and we add x for the products to note an increase in concentration. We're basically showing what happens during a using this table and describing the initial and final conditions.

You may be wondering what happens after filling in the ICE table. Now that we have solved for the concentrations, we can use our constant, K, to solve for x.

Solving for K with Equilibrium Concentrations

Finally, we can plug these concentrations into our formula for the constant to solve:

K = [CH₃COO⁻][H⁺] / [CH₃COOH]

K was given as 1.8 * 10⁻⁵ and we can plug the values in the "" row of the ICE table in:

K = 1.8 * 10⁻⁵ = [x][x] / [1-x] = x² / 1-x

From here, we could multiply both sides by (1-x) and use the quadratic formula, but remember that our K value is super super small. When K is small, we aren’t going to have produced much product at meaning that 1 - x is incredibly close to 1. We’ll elaborate a bit on this later, but for this reason, we can approximate 1 - x to be 1 to make our calculations easier.

Note that 1-x is a tiiiny bit less than 1 in reality, but because x << 1 (x is much less than 1), we can make the approximation. It basically makes such a little difference in our calculations that we can eliminate it. Therefore, we can write the following equation to solve for x:

x² / 1 = 1.8 * 10⁻⁵

x = √(1.8 * 10⁻⁵) = 0.0042.

Yay! We solved for x. Now we know that the concentrations of [CH₃COO⁻] and of [H+] are 0.0042 (because they’re x according to our ICE table). Similarly, we can say that the concentration of [CH₃COOH] is 1 - 0.0042, which is 0.9958. This is pretty close to 1, which makes sense because we’re not going to lose too much reactant to form products at .Note that the units in ICE boxes will typically be Molarity to represent concentration.

When to Use ICE Tables

This strategy can be applied to any you’re given as long as you have:

1. The constant

2. Some (s) (you don’t always start at the beginning, but any ungiven concentrations will be 0)

You can also use ICE tables to solve for initial concentrations, and in turn, Q. There are a ton of calculations that get unlocked once you learn this technique, but we'll get into that later in this unit and in unit eight. Unit eight is all about acids and bases, and ICE tables are particularly useful when working with reactions that involve a weak acid or weak base.

5% Approximation: Eliminating X

Many students struggle with the idea of simply dropping x from the denominator when using an . Let’s think about this idea completely isolated from ICE Boxes or even . Let’s suppose we have a number that is super tiny, like, astronomically tiny. Let’s set a number x = 0.0000001.

If we want to find a value like 3+x, we could manually add 3 and 0.0000001 to get 3.0000001, but note that this value is basically 3. Sure, it’s slightly off but when we’re doing calculations like x²/(3+x) = 1.43 * 10⁻³ (these are just made-up numbers), being able to approximate 3+x to 3 makes our calculations way easier. Typically, scientists say that we can make this approximation when it does not lead to an error larger than 5%, which is why it's called the 5% approximation or 5% rule.

However, on the AP Chemistry exam, you can almost always make this approximation and it makes your concentration calculations worlds easier because you won’t have to deal with a quadratic. The one approximation you want to be careful not to make is approximations that directly affect x. For example, even though it’s technically valid as an approximation, saying “4x = 0” would screw up calculations way more because they would always equal 0 which ruins the point of finding concentrations!

Basically, in any problem (as far as AP Chemistry is concerned), we can approximate a + x or a - x to be just a because x is much much less than a.

Practicing with ICE Boxes

Consider the following :

H₂CO₃ ⇌ HCO₃⁻ + H⁺ (K = 4.3 x 10⁻⁷)

At , what is the concentration of [HCO₃⁻] if the began with an initial H₂CO₃ concentration of 1.2 M?

Let’s write out an to solve this problem!

Note that the stoichiometric coefficients are one in this as well, so there is no number in front of the x in the change row of the ICE table. Say the of H₂CO₃ was 2, how would that affect the above ICE table? Well, the "change" for H₂CO₃ would be -2x instead of just -x. Now, let's get back to the math for this problem and solve for the concentration of HCO₃⁻.

4.3 * 10⁻⁷ = [x][x] / [1.2 - x] ≈ x² / 1.2 (this step is using the 5% rule)

x² = 1.2 * (4.3 * 10⁻⁷) = 5.16 * 10⁻⁷

[HCO₃⁻] = x = √(5.16 * 10⁻⁷) = 0.0007.