For X Y, K = 0.50, initial [X] = 1.0 M and no Y. Set up the equation in x and solve. [3 points]

United StatesChemistrySyllabus dot point

How do we calculate the equilibrium concentrations of all species from initial concentrations and the value of K?

Topic 7.7 Calculating Equilibrium Concentrations: use an ICE table and the value of K to calculate equilibrium concentrations, including the use of the small-x (5%) approximation where valid.

A focused answer to AP Chemistry Topic 7.7, covering using an ICE table with a known K to solve for equilibrium concentrations, setting up and solving the resulting equation, and the small-x approximation, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Setting up the calculation
The small-x approximation
Solving the full equation
Try this

What this topic is asking

The College Board (Topic 7.7) wants you to use an ICE table together with the value of K to calculate equilibrium concentrations, including the small-x (5%) approximation where it is valid. This is the inverse of Topic 7.4: there you found $K$ from concentrations; here you find concentrations from $K$ .

Setting up the calculation

The structure mirrors Topic 7.4, but now $K$ is known and $x$ is the unknown. The equilibrium expression becomes an equation: products over reactants (in terms of $x$ ) equals $K$ . Depending on the stoichiometry, this may be linear, quadratic or higher.

The small-x approximation

A small $K$ means the reaction barely proceeds, so $x$ is tiny relative to the starting amount and dropping it changes the answer negligibly. After solving, you must check: compute $\dfrac{x}{C_0} \times 100\%$ and confirm it is below 5%. If it exceeds 5%, the approximation is not valid and you solve the full quadratic.

Solving the full equation

When the approximation fails, or when the equation is quadratic by structure, solve it exactly. For an expression like $\dfrac{x^2}{C_0 - x} = K$ , multiply out to get $x^2 + Kx - KC_0 = 0$ and apply the quadratic formula, taking the physically meaningful (positive, smaller than $C_0$ ) root. Always discard roots that give negative concentrations.

Equilibrium concentrations with the small-x approximation

For $\text{A}(aq) \rightleftharpoons \text{B}(aq) + \text{C}(aq)$ , $K = 1.0 \times 10^{-4}$ , with initial $[\text{A}] = 0.20$ M and no products.

step 1 Build the ICE table

Initial: $\text{A}\ 0.20$ , $\text{B}\ 0$ , $\text{C}\ 0$ . Change: $\text{A}\ -x$ , $\text{B}\ +x$ , $\text{C}\ +x$ . Equilibrium: $\text{A}\ 0.20 - x$ , $\text{B}\ x$ , $\text{C}\ x$ .

step 2 Write the equilibrium equation

$K = \dfrac{x^2}{0.20 - x} = 1.0 \times 10^{-4}$ .

step 3 Apply the small-x approximation

Assume $0.20 - x \approx 0.20$ : $x^2 = (1.0 \times 10^{-4})(0.20) = 2.0 \times 10^{-5}$ , so $x = 4.5 \times 10^{-3}$ M.

step 4 Check validity

$\dfrac{4.5 \times 10^{-3}}{0.20} \times 100\% = 2.2\%$ , below 5%, so the approximation is valid. Equilibrium: $[\text{A}] \approx 0.20$ , $[\text{B}] = [\text{C}] = 4.5 \times 10^{-3}$ M.

Try this

Q1. For $\text{X} \rightleftharpoons \text{Y}$ , $K = 0.50$ , initial $[\text{X}] = 1.0$ M and no Y. Set up the equation in $x$ and solve. [3 points]

Cue. $K = \dfrac{x}{1.0 - x} = 0.50$ , so $x = 0.50(1.0 - x)$ , giving $x = 0.33$ M; $[\text{Y}] = 0.33$ , $[\text{X}] = 0.67$ M.

Q2. State the test that confirms the small-x approximation is valid. [1 point]

Cue. $x$ is less than 5% of the initial concentration.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For

\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)

K_a = 1.8 \times 10^{-5}

and the initial

[\text{HA}] = 0.100

M (no products initially). (a) Construct an ICE table in terms of

x

. (b) Write the equilibrium expression in terms of

x

. (c) Using the small-

x

approximation, calculate

[\text{H}^+]

. (d) Verify that the approximation is valid (less than 5%).

Show worked answer →

A 4-point quantitative FRQ with the small-x approximation.

(a) ICE table (1 point): Initial: $\text{HA}\ 0.100$ , $\text{H}^+\ 0$ , $\text{A}^-\ 0$ . Change: $\text{HA}\ -x$ , $\text{H}^+\ +x$ , $\text{A}^-\ +x$ . Equilibrium: $\text{HA}\ 0.100 - x$ , $\text{H}^+\ x$ , $\text{A}^-\ x$ .
(b) Expression (1 point): $K_a = \dfrac{x^2}{0.100 - x} = 1.8 \times 10^{-5}$ .
(c) Solve with approximation (1 point): assume $0.100 - x \approx 0.100$ ; then $x^2 = (1.8 \times 10^{-5})(0.100) = 1.8 \times 10^{-6}$ , so $x = 1.3 \times 10^{-3}$ M $= [\text{H}^+]$ .
(d) Validity (1 point): $\dfrac{1.3 \times 10^{-3}}{0.100} \times 100\% = 1.3\%$ , which is less than 5%, so the approximation is valid.

Markers reward the ICE table, the expression in $x$ , the approximate $[\text{H}^+]$ , and the 5% validity check.

AP 2021 (style)1 marksSection I (multiple choice). The small-

x

approximation

([\text{HA}]_0 - x \approx [\text{HA}]_0)

is valid when (A)

K

is large (B)

x

is small compared with the initial concentration (C) the reaction goes to completion (D) there are no products initially. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The approximation drops $x$ from the denominator, which is justified only when $x$ is small relative to the initial concentration (typically less than 5%). A small $K$ tends to make $x$ small, but the validity test is on $x$ itself. The trap is (A): a small $K$ (not a large one) supports the approximation.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)