U4.6 · Introduction to TitrationPractice · 滴定入门

KHP and NaOH react 1 : 1. n(KHP) = 0.5105 ÷ 204.22 = 0.002500 mol = n(NaOH), so [NaOH] = 0.002500 ÷ 0.02465 = 0.1014 M. All the data have 4 significant figures.

n(HCl) = 0.1000 × 0.03040 = 0.003040 mol. $\ce{Na2CO3}$ and HCl react 1 : 2, so n($\ce{Na2CO3}$) = 0.001520 mol and the molarity is 0.001520 ÷ 0.02500 = 0.06080 M. Ignoring the 1 : 2 ratio gives 0.1216 M.

The equivalence point is where the reactants have combined exactly in the ratio of the equation: HCl and NaOH react 1 : 1, so the moles of NaOH added equal the moles of HCl. Equal volumes hold only when the concentrations are equal; the color change marks the end point, which a good indicator puts close to, but not necessarily exactly at, the equivalence point; and titrating a weak acid gives a pH above 7 at equivalence.

Water left inside the pipet dilutes the acid drawn in, so the 25.00 mL delivered holds fewer moles of acid than it should. Less base is needed, and dividing by 25.00 mL as usual gives a concentration that is too low. If the acid is diluted to 0.95 of its strength, the result is about 5% low. The volume is indeed 25.00 mL, but it contains less acid.

n($\ce{MnO4-}$) = 0.02000 × 0.01840 = 3.680 × 10⁻⁴ mol. Each $\ce{MnO4-}$ gains 5 electrons and each $\ce{Fe^2+}$ loses 1, so they react 1 : 5: n($\ce{Fe^2+}$) = 0.001840 mol, and $[\ce{Fe^2+}]$ = 0.001840 ÷ 0.02500 = 0.07360 M. $\ce{MnO4-}$ is purple, so the first lasting pink color marks the end point without any added indicator.

The net ionic equation is Ba²⁺ + 2OH⁻ + 2H⁺ + SO₄²⁻ → BaSO₄(s) + 2H₂O. At first the solution holds Ba²⁺ and OH⁻ (about 26 mS/cm); as acid is added they are removed as precipitate and water, so the conductivity falls in a straight line. At the equivalence point (0.0500 × 25.0 = 0.100 × 12.5 mmol) almost no free ions remain. After that the extra H⁺ and SO₄²⁻ stay in solution and the conductivity rises again. NaCl solution is neutral yet conducts well, so “neutral means nonconducting” is wrong.

The equivalence point is in the middle of the steep rise: about 16.0 mL (pH 7 for a strong acid with a strong base). n(NaOH) = 0.100 × 0.0160 = 1.60 × 10⁻³ mol = n(HCl), so [HCl] = 1.60 × 10⁻³ ÷ 0.0200 L = 0.0800 M. Choosing 0.125 M flips the volume ratio (0.100 × 20.0 ÷ 16.0).