| English | 释义 | 拼音 |
|---|---|---|
| titre/ˈtɪtə/ | 滴定体积 | dī dìng tǐ jī |
Can a measured titre 滴定体积 reveal an unknown concentration?
- A titration finds how much of one solution reacts with a measured portion of another. Supplement calculations use the balanced equation, not an automatic 1 : 1 rule.
- A titre is the volume delivered from the burette: final reading minus initial reading. Use the experimental results specified in the question.
Convert readings to an amount
- Suppose a burette contains $0.100\ \dfrac{\mathrm{mol}}{\mathrm{dm^3}}$ HCl. Readings of 1.20 and 21.20 cm³ give a titre of 20.00 cm³ = 0.02000 dm³.
- The amount delivered is $n=cV=0.100\times0.02000=0.00200\ \mathrm{mol}$ HCl. Use the difference in readings, not the final reading alone.
A burette reading changes from 2.10 to 26.50 cm³. What titre was delivered, in cm³?
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Titre=26.50−2.10=24.40 cm³.
Find the amount in a 24.0 cm³ titre of 0.150 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$ HCl, in mol.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
n=cV=0.150×0.0240=0.00360 mol.
Use the balanced ratio before dividing by volume
- For HCl + NaOH → NaCl + H₂O, the ratio is 1 : 1. The acid therefore reacts with 0.00200 mol NaOH.
- If the flask contained 25.0 cm³ NaOH, its concentration was $c=n/V=0.00200/0.0250=0.0800\ \dfrac{\mathrm{mol}}{\mathrm{dm^3}}$. Divide by the volume of that original NaOH portion, not the combined acid-plus-alkali volume.
Handle a ratio that is not one to one
- In H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, one mole of acid reacts with two moles of alkali. Equal concentrations need not mean equal reacting volumes.
- If 10.0 cm³ of $0.200\ \dfrac{\mathrm{mol}}{\mathrm{dm^3}}$ H₂SO₄ neutralises 25.0 cm³ NaOH, acid amount $=0.200\times0.0100=0.00200\ \mathrm{mol}$. Alkali amount is 0.00400 mol, so its concentration is $0.00400/0.0250=0.160\ \dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
For titres 19.90, 20.00 and 20.10 cm³ that the question identifies as concordant, mean titre $=(19.90+20.00+20.10)/3=20.00\ \mathrm{cm^3}$. Use the stated selection rule; do not include a rough result or choose a value only because it matches the expected answer.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What amount of NaOH reacts with 0.0030 mol H₂SO₄, in mol?
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The ratio is 1:2, so NaOH amount=2×0.0030=0.0060 mol.
The question states that 24.60, 24.70 and 24.80 cm³ are concordant titres. Find their mean in cm³.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
Mean=(24.60+24.70+24.80)/3=24.70 cm³.
Separate dilution from reaction
- Adding distilled water to the flask after measuring the unknown portion changes its concentration in the flask but not its moles. The titre needed to react with those moles is unchanged, assuming no loss and the same valid endpoint.
- The indicator marks an endpoint; careful dropwise addition near that point reduces overshooting. If too much titrant is added, the recorded titre is too large and any concentration calculated from it can be biased.
Do not use $c_1V_1=c_2V_2$ without checking the equation. It applies directly to this reaction calculation only for a 1 : 1 mole ratio. For other ratios, calculate moles and apply the coefficients explicitly.
After measuring a 25.0 cm³ alkali portion into a flask, a student adds distilled water before titrating it with acid. No alkali is lost. Explain whether the acid titre needed for neutralisation changes.
Compare the reasoning and each criterion; equivalent wording is valid.
Present a complete chain of reasoning
- Readings → titre → dm³ → known moles → equation ratio → unknown moles → unknown concentration.
- Attach the unit and check the direction of any error against the actual calculation. The practical procedure is covered separately in the titration-technique lesson.
A titre measures reacting volume. The balanced equation determines the reacting mole ratio. The unknown concentration uses the original measured portion's volume.
15.0 cm³ of 0.200 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$ H₂SO₄ neutralises 20.0 cm³ NaOH. Use H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Find the NaOH concentration in $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
Acid n=0.200×0.0150=0.00300 mol. Alkali n=0.00600 mol. c=0.00600/0.0200=0.300 $\dfrac{\mathrm{mol}}{\mathrm{dm^3}}$.
Known acid is used to find an alkali concentration. The endpoint is overshot. How is the calculated alkali concentration affected?
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The extra acid inflates the inferred alkali amount, then the calculation divides by the same original alkali volume.