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Unit 7

7.7 Calculating Equilibrium Concentrations

5 min readjune 1, 2021

fiveable-dylan

Dylan Black


AP Chemistry 🧪

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7.7: Calculating Equilibrium Concentrations

One of the most important aspects of equilibrium is that it is the point at which a reaction “settles” and concentrations will not change because the rate of the forward reaction and reverse reaction will equal each other. However, what happens if we know the equilibrium constant for a reaction but don’t know what the equilibrium concentrations for a reaction are? Well, we have to do some math! In this section, you’ll learn how to use an ICE Box (sometimes called a RICE Box) to solve for equilibrium concentrations and learn some of the math behind one of AP Chemistry’s most important mathematical techniques.

ICE Boxes Explained

When solving for an equilibrium concentration, we use an ICE Box. The “I” stands for “initial”, the “C” stands for “change”, and the “E” stands for “equilibrium”. These boxes are used to show the initial concentrations, change in concentration, and the equilibrium concentrations for a reaction. Let’s take a look:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-CqDufCLEvU2l.png?alt=media&token=595c536d-6b5b-4ce9-995e-95734c2cc4ac

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Ok… just kidding. Here’s what a real ICE Box looks like with the reaction CH3COOH ⇌ CH3COO- + H+ (K = 1.8 * 10^(-5)):
Reaction
CH3COOH
CH3COO- 
H+
Initial
Change
Equilibrium
Let’s start filling this in! Let’s suppose we are starting with an initial CH3COOH concentration of 1M (initial concentrations are always given!). Because we’re starting at the very beginning before any reaction has occurred, [CH3COO-] and [H+] are 0:
Reaction
CH3COOH
CH3COO- 
H+
Initial
1M
0
0
Change
Equilibrium
Next, let’s think about how these quantities change. When this reaction 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 to denote this change. For reactants, we lose nx (where n is the stoichiometric coefficient), and for each product we gain nx. At equilibrium we add together the initial and change to get our final concentrations! Let’s see this on the table:
Reaction
CH3COOH
CH3COO- 
H+
Initial
1M
0
0
Change
-x
+x
+x
Equilibrium
1 - x
x
x
Finally, we can plug these equilibrium concentrations into our formula for the equilibrium constant to solve:
K = [CH3COO-][H+] / [CH3COOH] = [x][x] / [1-x] = x^2 / 1-x = 1.8 * 10^(-5).
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 equilibrium 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. 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. Therefore we can write the following equation to solve for x:
x^2 / 1 = 1.8 * 10^(-5)
x = √(1.8 * 10^(-5)) = 0.0042.
Therefore, we know that the equilibrium concentrations of [CH3COO-] and of [H+] are 0.0042 (because they’re x!). Similarly, we can say that the equilibrium concentration of [CH3COOH] is 1 - 0.0042 = 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 equilibrium. 
This strategy can be applied to any reaction you’re given as long as you have:
  1. The equilibrium constant
  2. Some initial concentration(s) (you don’t always start at the beginning, but any ungiven concentrations will be 0)

Wait… Why Can We Get Rid Of X?

Many students struggle with the idea of simply dropping x from the denominator when using an ICE box. Let’s think about this idea completely isolated from ICE Boxes or even equilibrium. 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 x^2/(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%
However, on AP Chemistry you can almost always make this approximation and it makes your equilibrium 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 are 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 equilibrium concentrations! 
Basically, in any ICE Box 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. If we have a K so large that we cannot, we  

Sample Problem

Consider the following reaction:
H2CO3 ⇌ HCO3- + H+ (K = 4.3 x 10^(-7))
At equilibrium, what is the concentration of [HCO3-] if the reaction began with an initial H2CO3 concentration of 1.2M?
Let’s write out an ICE Box to solve this problem!
Reaction
H2CO3
HCO3-
H+
Initial
1.2M
0M
0M
Change
-x
+x
+x
Equilibrium
1.2 - x
x
x
4.3 * 10^(-7) = [x][x] / [1.2 - x] ≈ x^2 / 1.2
x^2 = 1.2 * (4.3 * 10^(-7)) = 5.16 * 10^(-7)
[HCO3-] = x = √(5.16 * 10^(-7)) = 0.0007.

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