ARCH-C-16: Multi-Step Acid-Base Back-Titrations, Diprotic Ratios (H2SO4/Ca(OH)2), Thermometric Titration & Purity

Full 15-Q Timed Drill

Step 1 of 6 • The Core Idea

ARCH-C-16: Why the 160s Classroom Method Fails Under the 90s Clock

Standard Classroom Method (Too Slow)~160s (Blows 90s Clock)

First calculate the moles of

\text{NaOH}

25.0\text{ cm}^3

n(\text{NaOH}) = \frac{25.0}{1000} \times 0.16 = 0.0040\text{ mol}

. From the

1:2

stoichiometric ratio (

\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

n(\text{H}_2\text{SO}_4) = \frac{0.0040}{2} = 0.0020\text{ mol}

. Dividing by the acid volume

V = 20.0\text{ cm}^3 = 0.020\text{ dm}^3

gives

c(\text{H}_2\text{SO}_4) = \frac{0.0020}{0.020} = 0.10\text{ mol dm}^{-3}

(~95s).

The Fast ShortcutTarget: 45s (Saves 115s)

**Line 1 (Millimoles with

1:2

Ratio):**

n(\text{H}_2\text{SO}_4) = \frac{1}{2} \times 25.0 \times 0.16 = 2.0\text{ mmol}

.
**Line 2 (Divide by

20.0\text{ cm}^3

):**

c = \frac{2.0\text{ mmol}}{20.0\text{ cm}^3} = 0.10\text{ mol dm}^{-3}

(20s).

Where Students Lose Time: In 23 past UAT-UK questions (10.22% of Chemistry), candidates lose 115s by failing to balance h+ vs oh- equivalents: 1 mole of h2so4 or ca(oh)2 provides 2 equivalents of h+ or oh-. Instead, apply C9 with Maths 1 skill M3.5.

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