ACT Science: Titration and Acid-Base Chemistry Calculations

Titration is a lab technique where you add a solution of known concentration (the titrant) to a solution of unknown concentration (the analyte) until a chemical endpoint is reached. For acid-base titration, you add acid to a base (or vice versa) until pH reaches the equivalence point (where moles of acid equal moles of base). The key reaction: H⁺+OH⁻→H₂O. At equivalence, all H⁺ from acid is neutralized by all OH⁻ from base. On the ACT, you don't perform titrations; you interpret titration data or calculate unknown concentrations. Formula: M₁V₁=M₂V₂ (molarity×volume of acid = molarity×volume of base, assuming 1:1 ratio in the balanced equation). The equivalence point is when moles of acid added equals moles of base originally present.

Example: You titrate 25 mL of unknown NaOH with 0.1 M HCl. At equivalence, you've added 30 mL of HCl. Moles of HCl=0.1 M×0.030 L=0.003 mol. Since HCl:NaOH is 1:1, moles of NaOH=0.003 mol. Molarity of NaOH=0.003 mol/0.025 L=0.12 M.

Error 1: Forgetting to convert volume to liters before using M=moles/volume. 25 mL=0.025 L, not 25. Error 2: Ignoring the stoichiometry ratio if it's not 1:1. For H₂SO₄ (2 H⁺) with NaOH (1 OH⁻), the ratio is 1:2, so M₁V₁ must be multiplied by the ratio. Error 3: Confusing the analyte's unknown concentration with the titrant's known concentration. Titrant is known; analyte is unknown. Error 4: Reading the endpoint from a graph incorrectly (confusing equivalence point with half-equivalence or some other feature). Always check: (1) Which concentration is known? (2) What's the stoichiometry ratio? (3) Have I converted volumes to liters? (4) Am I solving for the right unknown?

Verification: After calculating the unknown molarity, check if it makes sense. Is it close to the titrant's molarity? Is it within typical lab ranges?

Scenario 1: 20 mL of 0.15 M HCl titrates 30 mL of unknown NaOH. Find molarity of NaOH. Using M₁V₁=M₂V₂: 0.15×20=M₂×30. M₂=0.1 M. Scenario 2: 25 mL of unknown H₂SO₄ is titrated with 40 mL of 0.2 M NaOH (ratio 1 H₂SO₄:2 NaOH). Find molarity of H₂SO₄. M₁V₁ (×2)=M₂V₂. M₁×25×2=0.2×40. M₁=0.16 M. Scenario 3: A graph shows pH vs mL of acid added, with equivalence point at 25 mL. 50 mL of 0.1 M base was titrated. Calculate moles of acid added. Moles=0.1 M×0.050 L=0.005 mol originally. At equivalence, 0.005 mol acid added. Volume of acid: 0.005 mol/(molarity) → need molarity from the graph or given separately. For each, label: unknown, known, stoichiometry ratio, calculation steps.

Daily drill: Solve one titration problem daily. Alternate between finding unknown molarity and calculating volume or moles. Practice converting units without errors.

Titration questions appear 1-2 times per ACT Science section and usually feel intimidating because of the vocabulary and graph interpretation. However, the math is straightforward (molarity, volume, moles) if you know the concept. Mastering titration gives you confidence on chemistry passages and unlocks points others avoid. This skill combines stoichiometry, molarity calculations, and graph reading—multiple skills at once—making it a high-value target for your preparation.

Spend 2-3 days on titration. Practice the calculation method, graph reading, and stoichiometry ratios. By test day, titration problems will feel manageable and you'll gain an advantage on chemistry sections.