Higher Tier, Chemistry-only: titration concentrations follow the equation ratio
導入済み| English | 意味 |
|---|---|
| titre/ˈtɪtə/ | 滴定体積 |
| end-point/end pɔɪnt/ | 滴定終点 |
What would explain this observation?
- Higher Tier: Equal volumes of acid and alkali do not necessarily react completely with each other. The concentrations and balanced coefficients both determine the required quantities.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- For reacting solutions, calculate known amount with n=cV, use the balanced coefficient ratio and divide the unknown amount by its volume. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O means one mole acid reacts with two moles alkali. A 1:1 shortcut is valid for HCl and NaOH, but not for this sulfuric-acid equation.
- titre 滴定体积: The volume delivered from a burette to reach the stated end-point 滴定终点; end-point: The observed change used to decide when to stop a titration.
For H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, how is acid amount converted to NaOH amount?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The equation H2SO4 + 2NaOH shows 1 mole of acid reacting with 2 moles of alkali, so NaOH moles = 2 x acid moles. Dividing grams by two ignores molar masses, and equal moles contradict the equation.
Match each technical term to its precise meaning.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- A pipette supplies the measured sample volume and a burette delivers the titre. The titre is final minus initial burette reading, not the final reading alone. Use consistent close results under the school’s stated concordance rule and exclude a rough trial from the accurate mean. A suitable indicator end-point approximates the required neutralisation condition; it is not evidence that every solution must be pH 7.
- Perform titration only as an actual supervised school task using approved dilute solutions, eye protection and the specified pipette filler. Rinse and use the apparatus as instructed, remove the filling funnel before readings and add dropwise near the end-point. Record readings at appropriate precision. These calculation exercises do not count as completing Required Practical 2, whose real method belongs in chemical changes.
Which two habits make the investigation or model in this case more defensible?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: 25.0 cm³ NaOH requires 20.0 cm³ of 0.100 $\dfrac{\text{mol}}{\text{dm}^3}$ H₂SO₄. Acid amount=0.100×0.0200=0.00200 mol. Alkali amount=2×0.00200=0.00400 mol. NaOH concentration=0.00400/0.0250=0.160 $\dfrac{\text{mol}}{\text{dm}^3}$. The ratio factor comes from the equation, not from the two measured volumes.
25.0 cm³ NaOH reacts with 15.0 cm³ of 0.200 $\dfrac{\text{mol}}{\text{dm}^3}$ H₂SO₄. Using the stated 1:2 ratio, find NaOH concentration. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
25.0 cm³ NaOH reacts with 15.0 cm³ of 0.200 mol per dm³ H₂SO₄. Using the stated 1:2 ratio, find NaOH concentration.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The result is 0.24 mol per dm³. Known: 15.0 cm^3 of 0.200 mol per dm^3 H2SO4; ratio 1:2. Acid moles = 0.200 x 0.0150 = 0.00300 mol. NaOH moles = 2 x 0.00300 = 0.00600 mol. Concentration = 0.00600/0.0250 = 0.24 mol per dm^3, using the NaOH volume 25.0 cm^3 = 0.0250 dm^3.
Check the conclusion and its limits
- Use the correct unknown volume in the final division. Mean titres must be selected from appropriate trials; averaging a rough result can spoil an accurate calculation. Keep practical skill evidence separate from written calculations and do not invent measurements as if they were collected by students.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
The final burette reading alone always equals the titre. This claim is false: The titre is the volume DELIVERED: final reading minus initial reading. A burette starting at 2.00 cm^3 and ending at 24.60 cm^3 delivered 22.60 cm^3 - not 24.60.
Higher Tier, Chemistry-only: titration concentrations follow the equation ratio: A pipette supplies the measured sample volume and a burette delivers the titre. The titre is final minus initial burette reading, not the final reading alone. Use consistent close results under the school’s stated concordance rule and exclude a rough trial from the accurate mean. A suitable indicator end-point approximates the required neutralisation condition; it is not evidence that every solution must be pH 7.
The final burette reading alone always equals the titre.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The titre is the volume DELIVERED: final reading minus initial reading. A burette starting at 2.00 cm^3 and ending at 24.60 cm^3 delivered 22.60 cm^3 - not 24.60.
The volume delivered from a burette to reach the stated end-point: write the technical term.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
titre means The volume delivered from a burette to reach the stated end-point.