Weak acids, buffers and titration decisions
引入| English | 释义 | 拼音 |
|---|---|---|
| dissociation | 解离 | jiě lí |
| buffer/ˈbʌfə/ | 缓冲溶液 | huǎn chōng róng yè |
| conjugate base/ˈkɒndʒuːɡeɪt beɪs/ | 共轭碱 | gòng è jiǎn |
| equivalence/ɪˈkwɪvələns/ | 等当点 | děng dāng diǎn |
Which pH model applies?
- A weak acid and a strong acid of the same concentration need not have the same pH. Strength concerns dissociation 解离, not the amount dissolved.
- Identify the acid/base species and any prior neutralisation before choosing a formula.
Use the weak-acid approximation
- For HA, $K_a=[H^+][A^-]/[HA]$. With small dissociation and negligible water contribution, $[H^+]\approx\sqrt{K_ac}$.
- For $K_a=1.80\times10^{-5}\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$ and $c=0.100\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$, $[H^+]\approx0.00134\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$ and pH ≈ 2.87. Dissociation is $(0.00134/0.100)\times100=1.34\%$, supporting the approximation.
A dilute monoprotic strong acid has [H⁺] = 0.0100 in moles per cubic decimetre. Find pH.
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pH = −log₁₀(0.0100) = 2.00.
Buffers 缓冲溶液 react with added reagent
- A buffer contains substantial weak acid HA and conjugate base 共轭碱 A⁻. Added H⁺ consumes A⁻; added OH⁻ consumes HA.
- First react the moles. Adding 0.00400 mol OH⁻ to 0.0100 mol HA leaves 0.00600 mol HA and makes 0.00400 mol A⁻.
At 298 K, $[OH^-]=0.0100\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$ gives $[H^+]=K_w/[OH^-]=10^{-12}\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$, so pH = 12.00 using $K_w=10^{-14}\,\dfrac{\mathrm{mol}^2}{\mathrm{dm}^6}$.
0.0100 mol HA receives 0.00300 mol OH⁻. What is the remaining HA:A⁻ ratio?
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Neutralisation consumes 0.00300 mol HA and forms that amount of A⁻, leaving 0.00700 mol HA.
The conjugate base component of an acid buffer consumes added _____ ions.
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A⁻ + H⁺ → HA consumes added hydrogen ions.
Use the remaining ratio
- For the buffer, $[H^+]\approx K_a[HA]/[A^-]$. The common final volume cancels when using the remaining mole ratio.
- With $K_a=1.80\times10^{-5}\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$ and HA:A⁻ = 6:4, $[H^+]=2.70\times10^{-5}\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$ and pH = 4.57. Both components must remain.
Read the curve before choosing an indicator
- At half-neutralisation of weak acid by strong base, pH = pKa. At equivalence 等当点, the reacting amounts match the balanced equation; pH need not be 7.
- An indicator transition must lie within the steep part of the curve. Beyond equivalence, excess strong reagent controls pH and the acid-buffer formula fails.
“A weak acid is always dilute” is false. Concentration and extent of dissociation describe different properties.
Do stoichiometry before equilibrium. Check whether the proposed formula still has the species it requires.
Every acid–base equivalence point must have pH 7.
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Weak-acid/strong-base equivalence is above 7 at 298 K; strong-acid/weak-base is below 7.
Which two conditions justify an indicator choice?
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The endpoint must respond to a small addition near equivalence. Starting pH does not determine a suitable transition.