Difficulty: Intermediate | Prerequisites: Part 1 study notes (pH, pOH, acid-base definitions), comfort with logarithms and solving quadratic-style expressions.

ICE tables are the workhorse tool for any equilibrium calculation in general chemistry. Once you move past strong acids and bases (which dissociate completely and need only a simple -log calculation), every pH problem involving a weak acid, a weak base, or a hydrolysing salt requires you to set up an Initial-Change-Equilibrium table, write the equilibrium expression, and solve for the unknown concentration. This is also the foundation for buffer calculations, titration curve analysis, and solubility problems later in the course. If you can set up and solve an ICE table confidently, you can handle most of what the rest of the semester will throw at you.

TL;DR

An ICE table tracks what happens to concentrations as a weak acid or base reaches equilibrium. You write the balanced equation, fill in Initial concentrations (what you start with), Change (-x for reactants, +x for products), and Equilibrium expressions (initial +/- change). Then you plug the equilibrium row into Ka or Kb, solve for x, and convert to pH. The small-x approximation (dropping x from the denominator) saves time when the acid or base is weak enough.

Key Terms

ICE table

A structured method for tracking Initial, Change, and Equilibrium concentrations for each species in a reversible reaction. Think of it as a bookkeeping grid that keeps your algebra from going sideways.

Initial concentration

The molarity of each species before any reaction takes place. Products of a weak acid or weak base dissociation start at 0.

Change (x)

The amount (in molarity) that shifts as the system moves toward equilibrium. For a weak acid HA dissociating into H+ and A-, HA decreases by x while H+ and A- each increase by x.

Equilibrium concentration

The final concentrations once the system has reached equilibrium: (initial concentration) minus or plus the change.

Small-x approximation

When Ka (or Kb) is very small relative to the initial concentration, x is negligible compared to the initial value, so you can simplify (C - x) to just C. This avoids the quadratic formula. A common rule of thumb: the approximation holds if x is less than 5% of the initial concentration.

Conjugate base hydrolysis

The reaction of a conjugate base (like ClO- or CN-) with water to produce OH- and the parent weak acid. This is why salts of weak acids dissolved in water are basic.

Henderson-Hasselbalch equation

A rearranged form of the Ka expression: pH = pKa + log([A-]/[HA]). Useful when you already know the ratio of conjugate base to acid, as in buffer calculations and Part 2 of this lab.

How to Set Up an ICE Table, Step by Step

1. Write the balanced equilibrium equation. For a weak acid: HA <-> H+ + A-. For a weak base reacting with water: B + H2O <-> BH+ + OH-.

1. Fill in the Initial row. The weak acid or base goes in at its given molarity. Products start at 0 (assuming no common-ion effect).

1. Fill in the Change row. The reactant decreases by x. Each product increases by x (or by a stoichiometric multiple of x, if the coefficients differ from 1).

1. Write the Equilibrium row. Each entry is the initial value plus the change: for HA it is (C - x), for H+ and A- it is x.

1. Write the equilibrium expression. Ka = [H+][A-] / [HA] = x^2 / (C - x).

1. Apply the small-x approximation if appropriate: if Ka / C < 0.05, you can replace (C - x) with C, giving x = sqrt(Ka x C). If x turns out to be more than 5% of C, go back and solve the full quadratic.

1. Convert x to pH: pH = -log(x) for a weak acid. For a weak base, x = [OH-], so pOH = -log(x) and pH = 14 - pOH.

Weak Acid Ka Calculations: Worked Examples

Hypochlorous Acid (HClO), 0.00128 M, Ka = 3.5 x 10^-8

Equation: HClO <-> H+ + ClO-

Ka = x^2 / (0.00128 - x). Since Ka is very small, approximate: x = sqrt(3.5 x 10^-8 x 0.00128) = 6.7 x 10^-6 M. Check: 6.7 x 10^-6 / 0.00128 = 0.5%, well under 5%. pH = -log(6.7 x 10^-6) = 5.17.

Carbonic Acid (H2CO3), 0.00144 M, Ka1 = 4.3 x 10^-7

Equation (first dissociation only): H2CO3 <-> H+ + HCO3-

Ka = x^2 / 0.00144 = 4.3 x 10^-7. x = sqrt(4.3 x 10^-7 x 0.00144) = 2.5 x 10^-5 M. pH = -log(2.5 x 10^-5) = 4.6.

Phosphoric Acid (H3PO4), 0.20 M, Ka1 = 7.5 x 10^-3

Equation (first dissociation): H3PO4 <-> H+ + H2PO4-

Ka = x^2 / 0.20 = 7.5 x 10^-3. x = sqrt(7.5 x 10^-3 x 0.20) = 3.9 x 10^-2 M. Check: 3.9 x 10^-2 / 0.20 = 19.5%, which exceeds 5%. Strictly, you should solve the quadratic here, but the lab uses this approximation and gets pH = -log(3.9 x 10^-2) = 1.4.

Lactic Acid (C3H6O3), 0.045 M, pH = 2.65 (Ka unknown, solve for it)

Working backwards: [H3O+] = 10^-2.65 = 0.0022 M. [HA]eq = 0.045 - 0.0022 = 0.0428 M. Ka = (0.0022)^2 / 0.0428 = 1.13 x 10^-4.

Weak Base Kb Calculations: Worked Examples

Ammonia (NH3), 0.00120 M

Equation: NH3 + H2O <-> NH4+ + OH-

Kb = Kw / Ka = (1.0 x 10^-14) / (5.6 x 10^-10) = 1.8 x 10^-5. Then Kb = x^2 / 0.00120, giving x = 1.46 x 10^-4 M. pOH = -log(1.46 x 10^-4) = 3.83. pH = 14 - 3.83 = 10.2.

Note: the Ka used here (5.6 x 10^-10) is the Ka of the conjugate acid NH4+, not of ammonia itself. This is a common source of confusion.

Cyanide Ion (CN-) from Sodium Cyanide, 0.0510 M

Equation: CN- + H2O <-> HCN + OH-

Kb = Kw / Ka(HCN) = (1.0 x 10^-14) / (6.2 x 10^-10) = 1.6 x 10^-5. x = sqrt(1.6 x 10^-5 x 0.0510) = 9 x 10^-4 M. pOH = -log(9 x 10^-4) = 3.05. pH = 14 - 3.05 = 10.95.

Salt Hydrolysis: Acidic and Basic Salts

When a salt dissolves in water, its ions may react with water (hydrolyse). Whether the solution ends up acidic or basic depends on the relative strengths of the parent acid and base.

Ammonium Chloride (NH4Cl), 0.00142 M, Acidic Salt

NH4+ is the conjugate acid of the weak base NH3. Cl- comes from the strong acid HCl and does not hydrolyse.

Equation: NH4+ + H2O <-> NH3 + H3O+

Ka(NH4+) = 5.6 x 10^-10. x = sqrt(5.6 x 10^-10 x 0.00142) = 8.9 x 10^-7 M. pH = -log(8.9 x 10^-7) = 6.05. The solution is slightly acidic, as expected.

Sodium Hypochlorite (NaClO), 0.00125 M, Basic Salt

ClO- is the conjugate base of the weak acid HClO. Na+ comes from the strong base NaOH and does not hydrolyse.

Equation: ClO- + H2O <-> HClO + OH-

Kb = Kw / Ka(HClO) = (1.0 x 10^-14) / (3.5 x 10^-8) = 2.9 x 10^-7. x = sqrt(2.9 x 10^-7 x 0.00125) = 1.9 x 10^-5 M. pOH = 4.72. pH = 9.28. The solution is basic.

Sodium Hydrogen Carbonate (NaHCO3), 0.35 M, Amphoteric Salt

HCO3- can act as both an acid (donating H+ to form CO3^2-) and a base (accepting H+ to form H2CO3). For an amphoteric species, the pH is calculated using the average of the two pKa values:

pH = (pKa1 + pKa2) / 2 = (-log(4.3 x 10^-7) + -log(5.6 x 10^-11)) / 2 = (6.37 + 10.25) / 2 = 8.31.

Note that the concentration of NaHCO3 does not appear in this formula. The pH of an amphoteric salt depends only on the Ka values of the parent acid.

Graphical Determination of Ka (Part 2 of the Lab)

Part 2 of the lab demonstrates how to determine Ka by measuring the pH of a weak acid solution at different points during a titration with NaOH. Rather than relying on a single measurement, you take several data points and use them together.

The Method

You start with a known concentration of weak acid (0.05 M in 100 mL) and add measured volumes of 0.200 M NaOH. At each point:

• Calculate [H3O+] from the measured pH: [H3O+] = 10^(-pH)

• Calculate moles of OH- added: n(OH-) = M(NaOH) x V(NaOH)

• That many moles of HA have been converted to A-, so n(A-) = n(OH-)

• [A-] = n(A-) / total volume

• [HA] = initial moles of HA minus n(A-), divided by total volume

• Ka = [H3O+] x [A-] / [HA]

Lab Data

All three solutions give Ka values close to 1.5 x 10^-4, which is consistent and confirms the acid identity.

The Graph

Plotting 1/[A-] on the y-axis against [H3O+] on the x-axis gives a straight line. The linear relationship (f(x) = 126907.8x + 21.46, R^2 = 1) confirms that the Ka expression holds across the measurements. The slope of this line relates to Ka: since Ka = [H3O+][A-]/[HA], rearranging gives a linear form where the slope and intercept can be used to extract Ka and the initial acid concentration.

The advantage of this graphical method over a single pH measurement is that it averages out experimental error across multiple data points, giving a more reliable Ka value.

Common Misconceptions

• Students sometimes set up ICE tables for strong acids like HCl or HNO3. There is no equilibrium to track: strong acids dissociate completely, so [H3O+] simply equals the initial acid concentration.

• Forgetting to use Kb (not Ka) when calculating the pH of a conjugate base in water. If you are given Ka for HClO and asked for the pH of NaClO, you must first convert: Kb = Kw / Ka.

• Mixing up which equilibrium constant to use for salt hydrolysis. For the cation of a weak base (like NH4+), use Ka. For the anion of a weak acid (like ClO- or CN-), use Kb.

• Blindly applying the small-x approximation without checking. Phosphoric acid at 0.20 M with Ka1 = 7.5 x 10^-3 gives x / C = 19.5%, which violates the 5% rule. The lab proceeds with the approximation anyway, but an exam may penalise you for not solving the quadratic.

Why It Matters / Exam Flags

• ICE table problems are near-guaranteed on any general chemistry exam covering equilibrium. Know the setup cold.

• Be prepared to calculate Ka from a given pH and concentration (the lactic acid problem is a classic exam format).

• The Kb = Kw / Ka conversion is tested frequently, especially in salt hydrolysis and conjugate base problems.

• Know when the small-x approximation fails and be ready to solve the quadratic. If your exam is multiple choice, the approximation almost always works. If it is free response, check the 5% rule and show your work.

• The graphical determination of Ka (plotting to get a straight line and extracting the constant from the slope) is a higher-level question that may appear on lab practical exams.

• Amphoteric species (like HCO3-) and the shortcut pH = (pKa1 + pKa2) / 2 are easy marks if you know the formula, and easy to lose if you try to set up a full ICE table instead.

Quick Self-Test

1. True or false: an ICE table is needed to find the pH of 0.01 M NaOH. (False. NaOH is a strong base; pOH = -log(0.01) = 2, pH = 12.)

1. Fill in the blank: to find Kb for the conjugate base of a weak acid, use Kb = _____ / Ka. (Kw, or 1.0 x 10^-14)

1. True or false: if Ka = 1.8 x 10^-5 and the initial acid concentration is 0.10 M, the small-x approximation is valid. (True. x = sqrt(1.8 x 10^-5 x 0.10) = 1.3 x 10^-3; 1.3 x 10^-3 / 0.10 = 1.3%, under 5%.)

1. Fill in the blank: for the amphoteric ion HCO3-, pH = (_____ + _____) / 2. (pKa1 + pKa2)

1. True or false: a solution of NaCN is acidic because CN- is a weak species. (False. CN- is the conjugate base of the weak acid HCN, so it hydrolyses to produce OH-, making the solution basic.)

Practice Q&A

Q: Set up an ICE table and calculate the pH of a 0.025 M solution of formic acid (HCOOH), Ka = 1.8 x 10^-4.

A: HCOOH <-> H+ + HCOO-. Initial: [HCOOH] = 0.025, [H+] = 0, [HCOO-] = 0. Change: -x, +x, +x. Equilibrium: 0.025 - x, x, x. Ka = x^2 / 0.025 = 1.8 x 10^-4. x = sqrt(4.5 x 10^-6) = 2.1 x 10^-3 M. Check: 2.1 x 10^-3 / 0.025 = 8.4%, which exceeds 5%, so solving the quadratic would be more accurate. Using the approximation: pH = -log(2.1 x 10^-3) = 2.68.

Q: A 0.050 M solution of sodium fluoride (NaF) has Ka(HF) = 6.8 x 10^-4. Calculate the pH.

A: F- is the conjugate base of HF. Kb = Kw / Ka = (1.0 x 10^-14) / (6.8 x 10^-4) = 1.47 x 10^-11. ICE table for F- + H2O <-> HF + OH-: x = sqrt(1.47 x 10^-11 x 0.050) = 8.6 x 10^-7 M. pOH = -log(8.6 x 10^-7) = 6.07. pH = 14 - 6.07 = 7.93. Slightly basic, as expected for the salt of a strong base and a weak acid.

Q: Explain why NH4Cl dissolves to give a solution with pH below 7.

A: NH4Cl dissociates into NH4+ and Cl-. Cl- comes from the strong acid HCl and does not react with water. NH4+ is the conjugate acid of the weak base NH3 and donates a proton to water: NH4+ + H2O <-> NH3 + H3O+. This generates extra H3O+, pushing the pH below 7.

Q: In the graphical Ka determination, why does plotting 1/[A-] vs [H3O+] give a straight line?

A: From Ka = [H3O+][A-]/[HA], rearranging to 1/[A-] = Ka / ([H3O+] x [HA]) shows a linear relationship between 1/[A-] and [H3O+] when the total acid concentration is constant. The slope relates to Ka and the initial concentration of the acid.

Connections to Other Topics

Buffer solutions are the direct next step: a buffer is a weak acid mixed with its conjugate base (or a weak base with its conjugate acid), and the Henderson-Hasselbalch equation is simply the Ka expression rearranged. Every ICE table skill transfers directly.

Titration curves plot pH against volume of titrant added. The half-equivalence point (where [HA] = [A-]) gives pH = pKa, a fact that falls straight out of the Ka expression. The graphical method in Part 2 of this lab is a simplified version of the same idea.

Solubility equilibria (Ksp) use the same ICE-table framework, just with slightly soluble salts instead of acids and bases. If you can set up Ka and Kb ICE tables, Ksp problems will feel familiar.

Related Terms / Search Tags

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