The best examples of ICE tables in equilibrium calculations

Starting with concrete examples of ICE tables in equilibrium calculations

Let’s skip the vague theory and go straight to what you actually use on homework and exams: examples of ICE tables in equilibrium calculations that mirror real problem types. Then we’ll unpack the patterns.

Example 1: Weak acid dissociation (classic ICE table pattern)

Consider acetic acid in water:

\[ \text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CH}_3\text{COO}^-(aq) \]

Suppose you have 0.10 M acetic acid and no acetate initially. The acid dissociation constant is \( K_a = 1.8 \times 10^{-5} \) at 25 °C.

Set up the ICE table in terms of molarity:

I (Initial)
\[ [\text{CH}_3\text{COOH}] = 0.10, \quad [\text{H}_3\text{O}^+] \approx 0, \quad [\text{CH}_3\text{COO}^-] = 0 \]

C (Change)
Let \(x\) be the amount of acid that dissociates:

\[ \Delta[\text{CH}_3\text{COOH}] = -x, \quad \Delta[\text{H}_3\text{O}^+] = +x, \quad \Delta[\text{CH}_3\text{COO}^-] = +x \]

E (Equilibrium)
\[ [\text{CH}_3\text{COOH}] = 0.10 - x \] \[ [\text{H}_3\text{O}^+] = x \]
\[ [\text{CH}_3\text{COO}^-] = x \]

Now plug into the equilibrium expression:

\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = \frac{x^2}{0.10 - x} = 1.8 \times 10^{-5} \]

Because \(K_a\) is small and the initial concentration is moderate, it’s reasonable to approximate \(0.10 - x \approx 0.10\):

\[ \frac{x^2}{0.10} \approx 1.8 \times 10^{-5} \Rightarrow x^2 \approx 1.8 \times 10^{-6} \Rightarrow x \approx 1.34 \times 10^{-3} \text{ M} \]

So the equilibrium \([\text{H}_3\text{O}^+] \approx 1.3 \times 10^{-3}\,\text{M}\), giving a pH around 2.87.

This is a textbook example of an ICE table in equilibrium calculations: weak acid, single unknown, small-\(x\) approximation.

Example 2: Weak base equilibrium – parallel to weak acid

Now flip the script with ammonia:

\[ \text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq) \]

Say you start with 0.25 M NH₃, and \(K_b = 1.8 \times 10^{-5}\). The ICE table is almost a mirror of the weak acid case, another one of the best examples of ICE tables in equilibrium calculations for students.

I:
\[ [\text{NH}_3] = 0.25, \quad [\text{NH}_4^+] = 0, \quad [\text{OH}^-] \approx 0 \]

C:
\[ -x, +x, +x \]

E:
\[ [\text{NH}_3] = 0.25 - x, \quad [\text{NH}_4^+] = x, \quad [\text{OH}^-] = x \]

\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = \frac{x^2}{0.25 - x} = 1.8 \times 10^{-5} \]

Again, \(0.25 - x \approx 0.25\):

\[ x^2 \approx (1.8 \times 10^{-5})(0.25) = 4.5 \times 10^{-6} \Rightarrow x \approx 2.1 \times 10^{-3} \text{ M} \]

Now \([\text{OH}^-] = 2.1 \times 10^{-3}\,\text{M}\), so \(pOH \approx 2.68\) and \(pH \approx 11.32\).

This is a clean example of an ICE table in equilibrium calculations where you can see the symmetry between weak acids and weak bases.

For reference, updated equilibrium constants for common acids and bases are tabulated by many university chemistry departments; for example, see the University of Texas at Austin’s general chemistry data tables: https://web.chem.utexas.edu/chemdex/

Example 3: Gas-phase equilibrium with partial pressures

Next, a gas-phase reaction that shows up constantly in general chemistry:

\[ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g) \]

Suppose you start with 0.500 atm of N₂O₄ and no NO₂ in a sealed container at a temperature where \(K_p = 0.15\).

I:
\(P_{\text{N}_2\text{O}_4} = 0.500\,\text{atm}, \quad P_{\text{NO}_2} = 0\)

C:
Let \(x\) be the change in pressure of N₂O₄ that dissociates:

\[ \Delta P_{\text{N}_2\text{O}_4} = -x, \quad \Delta P_{\text{NO}_2} = +2x \]

E:
\[ P_{\text{N}_2\text{O}_4} = 0.500 - x \] \[ P_{\text{NO}_2} = 2x \]

Now apply the equilibrium expression:

\[ K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} = \frac{(2x)^2}{0.500 - x} = 0.15 \]

\[ \frac{4x^2}{0.500 - x} = 0.15 \Rightarrow 4x^2 = 0.15(0.500 - x) \]
\[ 4x^2 = 0.075 - 0.15x \Rightarrow 4x^2 + 0.15x - 0.075 = 0 \]

Solve the quadratic (use the positive root):

\[ x \approx 0.12 \text{ atm} \]

So:

\[ P_{\text{N}_2\text{O}_4} \approx 0.500 - 0.12 = 0.38\,\text{atm} \] \[ P_{\text{NO}_2} \approx 2(0.12) = 0.24\,\text{atm} \]

This is one of the best examples of ICE tables in equilibrium calculations using partial pressures instead of concentrations.

Example 4: Solubility product (Ksp) and ICE tables

ICE tables are not just for gases and acid–base systems. They’re also powerful for slightly soluble salts.

Take calcium fluoride, \(\text{CaF}_2(s)\), with:

\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq) \]

Let \(K_{sp} = 3.9 \times 10^{-11}\) (a typical textbook value at 25 °C). You want the molar solubility in pure water.

I:
Before any dissolving, concentrations are effectively zero:

\[ [\text{Ca}^{2+}] = 0, \quad [\text{F}^-] = 0 \]

C:
Let \(s\) be the molar solubility of CaF₂:

\[ \Delta[\text{Ca}^{2+}] = +s, \quad \Delta[\text{F}^-] = +2s \]

E:
\[ [\text{Ca}^{2+}] = s, \quad [\text{F}^-] = 2s \]

Now:

\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s(2s)^2 = 4s^3 = 3.9 \times 10^{-11} \]

\[ s^3 = \frac{3.9 \times 10^{-11}}{4} = 9.75 \times 10^{-12} \Rightarrow s \approx 2.1 \times 10^{-4}\,\text{M} \]

This is a straightforward example of an ICE table in equilibrium calculations for solubility, and it generalizes nicely to more complicated salts.

For more on solubility equilibria and Ksp data, see the NIST Chemistry WebBook: https://webbook.nist.gov/chemistry/

Example 5: Common-ion effect with ICE tables

Now take that same CaF₂, but dissolve it in a solution that already contains fluoride, say 0.10 M NaF. The reaction is the same:

\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq) \]

I:
\[ [\text{Ca}^{2+}] = 0, \quad [\text{F}^-] = 0.10 \]

C:
Let \(s’\) be the additional solubility of CaF₂ in the presence of fluoride:

\[ \Delta[\text{Ca}^{2+}] = +s’, \quad \Delta[\text{F}^-] = +2s’ \]

E:
\[ [\text{Ca}^{2+}] = s’ \]
\[ [\text{F}^-] = 0.10 + 2s’ \]

Now apply \(K_{sp}\):

\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s’(0.10 + 2s’)^2 = 3.9 \times 10^{-11} \]

Because \(s’\) will be very small compared to 0.10 M, you can approximate:

\[ [\text{F}^-] \approx 0.10 \]

So:

\[ 3.9 \times 10^{-11} \approx s’(0.10)^2 = s’(0.010) \Rightarrow s’ \approx 3.9 \times 10^{-9}\,\text{M} \]

The solubility plummets from \(2.1 \times 10^{-4}\,\text{M}\) to \(3.9 \times 10^{-9}\,\text{M}\) in the presence of the common ion. This is one of the most instructive examples of ICE tables in equilibrium calculations because it shows how the same setup captures very different physical behavior.

Example 6: Buffer system and Henderson–Hasselbalch via ICE

Students often memorize the Henderson–Hasselbalch equation but forget where it comes from. An ICE table connects the dots.

Consider a buffer made from 0.20 M acetic acid and 0.20 M sodium acetate. The equilibrium is:

\[ \text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CH}_3\text{COO}^-(aq) \]

I:
\[ [\text{CH}_3\text{COOH}] = 0.20, \quad [\text{CH}_3\text{COO}^-] = 0.20, \quad [\text{H}_3\text{O}^+] \approx 0 \]

C:
Let \(x\) be the small amount of acid that dissociates:

\[ -x, +x, +x \]

E:
\[ [\text{CH}_3\text{COOH}] = 0.20 - x \] \[ [\text{CH}_3\text{COO}^-] = 0.20 + x \]
\[ [\text{H}_3\text{O}^+] = x \]

Plug into \(K_a\):

\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = \frac{x(0.20 + x)}{0.20 - x} \]

Because \(x\) is tiny compared with 0.20 M:

\[ K_a \approx x \cdot \frac{0.20}{0.20} = x \]

So \([\text{H}_3\text{O}^+] \approx K_a\), and the pH is near the \(pK_a\). This ICE-table-based reasoning is what underlies the Henderson–Hasselbalch equation used widely in biochemistry and physiology; for example, blood pH buffering with carbonic acid and bicarbonate is discussed in detail by the National Library of Medicine: https://www.ncbi.nlm.nih.gov/books/NBK541123/

This is a more applied example of an ICE table in equilibrium calculations, directly tied to real biological systems.

Example 7: Reaction quotient Q vs. K using an ICE-style setup

Sometimes you’re not at equilibrium yet. You’re asked whether a system will shift left or right. You can still use an ICE-style table to organize data.

Take:

\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g) \]

Suppose at a given temperature \(K_c = 50\). A mixture contains:

\[ [\text{H}_2] = 0.10\,\text{M}, \quad [\text{I}_2] = 0.10\,\text{M}, \quad [\text{HI}] = 0.10\,\text{M} \]

You can think of the present state as the “I” row and compute the reaction quotient:

\[ Q_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.10)^2}{(0.10)(0.10)} = 1 \]

Because \(Q_c = 1 < K_c = 50\), the system will shift right, forming more HI. To find the actual equilibrium concentrations, you then build a full ICE table with \(x\) as the change and solve.

This is a slightly different but very practical example of ICE tables in equilibrium calculations, because it shows how ICE logic and the reaction quotient \(Q\) work together.

How to recognize when an ICE table is the right tool

After seeing several real examples of ICE tables in equilibrium calculations, patterns start to appear:

• You have a reversible reaction with a known \(K\) (\(K_c\), \(K_p\), \(K_a\), \(K_b\), \(K_{sp}\)).
• You’re given initial amounts and asked for equilibrium amounts, or vice versa.
• Stoichiometric relationships between species matter, not just their absolute amounts.

Current general chemistry curricula (2024–2025) still lean heavily on ICE tables because they bridge stoichiometry, algebra, and conceptual understanding. If you look at recent open-access textbooks, like those hosted by LibreTexts (https://chem.libretexts.org), you’ll see ICE tables featured in almost every equilibrium chapter.

The best examples of ICE tables in equilibrium calculations usually fall into a few categories:

• Weak acids and bases (pH, pOH, buffer design)
• Gas-phase reactions (using \(K_p\) or \(K_c\))
• Solubility and precipitation (\(K_{sp}\), common-ion effect)
• Biological and environmental systems (buffered blood, ocean carbonate chemistry)

Once you can handle the examples above, you can adapt to nearly any equilibrium scenario you’re likely to see in an undergraduate course.

Common mistakes students make with ICE tables

Even with the best examples of ICE tables in equilibrium calculations in front of them, students often trip over the same issues:

Sign errors in the Change row
If a species is formed, its change is positive; if it’s consumed, it’s negative. Using the balanced chemical equation as your guide is non‑negotiable.

Forgetting stoichiometric coefficients
If the equation has a coefficient of 2 in front of a product, its change is 2x, not x. You saw this in the N₂O₄/NO₂ and CaF₂ examples.

Dropping or misusing approximations
The small-\(x\) shortcut (like \(0.10 - x \approx 0.10\)) only works when \(x\) is much smaller than the starting value. A quick rule of thumb many instructors use: if \(x\) ends up less than about 5% of the initial concentration, the approximation is usually acceptable.

Using the wrong K expression
Always write the equilibrium expression from scratch: products over reactants, raised to their coefficients, omitting pure solids and liquids.

These are all fixable with practice, and working through a variety of examples of ICE tables in equilibrium calculations is the fastest way to build that muscle memory.

FAQ: ICE tables and equilibrium

Q: Why do teachers keep using ICE tables when calculators can just solve the equations?
Because ICE tables force you to connect the balanced reaction, the stoichiometry, and the equilibrium expression. They show why the algebra looks the way it does. That conceptual link is what you need for exams, lab work, and later courses.

Q: Can you give another quick example of an ICE table in an equilibrium calculation?
Sure. Consider \(\text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g)\) with initial pressures 0.50 atm CO, 0.50 atm Cl₂, and 0 atm COCl₂, and \(K_p = 4.0\). Let \(x\) be the amount of COCl₂ formed: the Change row is \(-x, -x, +x\), and the Equilibrium row is \(0.50 - x, 0.50 - x, x\). Plug into \(K_p = \frac{x}{(0.50 - x)^2}\) and solve. It’s the same ICE logic you’ve already seen.

Q: Do ICE tables work for reactions with more than two or three species?
Yes, but the algebra gets more intense. You still track Initial, Change, and Equilibrium for each species. In practice, instructors often choose examples of ICE tables in equilibrium calculations with at most three unknowns so that the math stays manageable.

Q: Where can I find more practice problems and data for ICE table questions?
Many U.S. universities post free problem sets and data tables. Look at general chemistry resources from schools like MIT (https://ocw.mit.edu) or other .edu-hosted courses. They typically include multiple examples of ICE tables in equilibrium calculations with detailed solutions.

Q: Are ICE tables still emphasized in newer (2024–2025) chemistry curricula?
Yes. Even with more digital tools and simulation software in classrooms, ICE tables remain standard in AP Chemistry, general chemistry, and MCAT prep materials. They’re still one of the best ways to teach the structure behind equilibrium problems.