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Connect proton transfer, conjugate stability and Lewis roles

A proton-transfer prediction requires matching conjugate pairs and compatible equilibrium conventions.

Learning outcome: Vary the donor and product conjugate-acid constants separately. Write the two ionization equations whose quotient gives K, then distinguish proton transfer from a non-proton Lewis association.

Starting recap

Refresh proton equivalents, conjugate pairs, logarithmic dimensionless activity, analytical amount/volume, mass and charge balances, and bounded observational criteria.

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Try

HA pKa 4 and BH+ pKa 6 give K=10^(6−4)=100 for HA+B⇌A−+BH+.

Open the experiment

Predict

Does the stronger donor acid have the larger or smaller pKa?

Your prediction and reason

Experiment

Vary the donor and product conjugate-acid constants separately. Write the two ionization equations whose quotient gives K, then distinguish proton transfer from a non-proton Lewis association.

Compare conjugate pairs through a proton-transfer constant

Worked teaching model · synthetic values · Fictional HA+B⇌A−+BH+, with both pKa values defined in the same solvent, temperature and standard-state convention. K=Ka(HA)/Ka(BH+). Supplied thermodynamic constants, not an experiment or a rate prediction.

Donor acid pKa: 4 dimensionless Product conjugate-acid pKa: 6 dimensionless

Reset keeps your prediction, captures and practice history.

log10 forward K versus product pKa: solid

Axes rescale. A concentration scan, imposed-pH distribution and titration curve have different horizontal meanings. Read each axis and preserve the difference between an imposed condition and an inferred equilibrium.

All calculated plot coordinates

log10 forward K versus product pKa
product conjugate-acid pKa	log10 K
2	-2
2.06	-1.94
2.12	-1.88
2.18	-1.82
2.24	-1.76
2.3	-1.7
2.36	-1.64
2.42	-1.58
2.48	-1.52
2.54	-1.46
2.6	-1.4
2.66	-1.34
2.72	-1.28
2.78	-1.22
2.84	-1.16
2.9	-1.1
2.96	-1.04
3.02	-0.98
3.08	-0.92
3.14	-0.86
3.2	-0.8
3.26	-0.74
3.32	-0.68
3.38	-0.62
3.44	-0.56
3.5	-0.5
3.56	-0.44
3.62	-0.38
3.68	-0.32
3.74	-0.26
3.8	-0.2
3.86	-0.14
3.92	-0.08
3.98	-0.02
4.04	0.04
4.1	0.1
4.16	0.16
4.22	0.22
4.28	0.28
4.34	0.34
4.4	0.4
4.46	0.46
4.52	0.52
4.58	0.58
4.64	0.64
4.7	0.7
4.76	0.76
4.82	0.82
4.88	0.88
4.94	0.94
5	1
5.06	1.06
5.12	1.12
5.18	1.18
5.24	1.24
5.3	1.3
5.36	1.36
5.42	1.42
5.48	1.48
5.54	1.54
5.6	1.6
5.66	1.66
5.72	1.72
5.78	1.78
5.84	1.84
5.9	1.9
5.96	1.96
6.02	2.02
6.08	2.08
6.14	2.14
6.2	2.2
6.26	2.26
6.32	2.32
6.38	2.38
6.44	2.44
6.5	2.5
6.56	2.56
6.62	2.62
6.68	2.68
6.74	2.74
6.8	2.8
6.86	2.86
6.92	2.92
6.98	2.98
7.04	3.04
7.1	3.1
7.16	3.16
7.22	3.22
7.28	3.28
7.34	3.34
7.4	3.4
7.46	3.46
7.52	3.52
7.58	3.58
7.64	3.64
7.7	3.7
7.76	3.76
7.82	3.82
7.88	3.88
7.94	3.94
8	4

Calculated under the declared model
Quantity	Value	Unit
Donor acid constant	1.000000e-4	dimensionless
Product conjugate-acid constant	1.000000e-6	dimensionless
Forward proton-transfer constant	100	dimensionless
Reverse proton-transfer constant	0.01	dimensionless

The stated proton transfer favors products thermodynamically.

Combine HA⇌H++A− with the reverse of BH+⇌H++B; hydronium cancels and K=Ka(HA)/Ka(BH+). Conjugate partners differ by exactly one proton. A Lewis base donates an electron pair and a Lewis acid accepts it; proton transfer is one subset, not every Lewis reaction. K alone supplies neither reaction speed nor final conversion.

These calculated packets are not performed chemical measurements. Keep feasible inventories, activity conventions, fitting observations and independent checks distinct.

The stated proton transfer favors products thermodynamically.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Identify the pairs before comparing strengths

In HA+B⇌A−+BH+, HA donates a proton and B accepts it. HA/A− and BH+/B are conjugate pairs: each differs by exactly one proton. Comparing HA with BH+ compares the two possible donors, not a donor with an unrelated anion.

Combine HA⇌H++A− with the reverse of BH+⇌H++B. The proton activity cancels, giving K=Ka(HA)/Ka(BH+)=10^[pKa(BH+)−pKa(HA)]. Both constants must use the same solvent, temperature and compatible standard states. At pKa 4 and 6, K=100; this favors products but does not imply instantaneous or complete conversion at every inventory.

Connect structure to a constrained prediction

A conjugate base stabilized by delocalization or electron-withdrawing groups can support stronger proton donation, with other factors comparable. For related oxoacids, additional electronegative oxygen environments often stabilize the conjugate base. Across different atoms, bond strength, size, solvation and electronegativity can compete. State the structural comparison rather than treating one slogan as a universal ranking rule.

Distinguish Lewis roles

A Lewis base donates an electron pair; a Lewis acid accepts it. Proton acceptance uses a base lone pair, but Lewis association need not transfer a proton: a donor pair binding an electron-deficient center is a broader category. Amphiprotic water can be either proton donor or acceptor depending on its partner.

Transfer and check

For donor pKa 6 and product conjugate-acid pKa 8, K=100 again. This matches the earlier transfer constant without making either acid equally strong in absolute terms. Explain why the same pKa difference fixes a relative transfer tendency, while a final composition still needs material balances.

Source and evidence boundary

Conceptual reference checked 30 October 2026: OpenStax Chemistry 2e. Prose, numerical packets and diagrams are original. All chemistry is fictional or supplied; no substance handling, mixing, titration apparatus or health interpretation is requested. The ideal concentration approximation, fixed temperature and inert-ion assumptions are explicit, not measured calibration. Retain your charge/mass balance, prediction, changed-case check and limits.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

How could resonance, induction or solvent stabilization change a conjugate-base stability argument?

Your explanation and limitations

Continue

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Previous: Bound an inferred constant and test a changed inventory Next: Use the water constant instead of assuming neutral always means seven

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