Salt effect or mechanism? Reading a Brønsted–Bjerrum plot
A student follows S2O8^2− + 2 I− → 2 SO4^2− + I2 at 25 °C by iodometry, keeping the oxidant and iodide in excess so that a second-order constant k can be extracted. To test the Brønsted–Bjerrum equation she varies ionic strength with NaClO4 and obtains: I = 0.010, 0.025, 0.050, 0.100 M giving k = 0.046, 0.057, 0.071, 0.093 M^−1 s^−1. A plot of log k against √I/(1+√I) is linear with slope ≈ 2.0, which she reads as proof that the rate-determining step is the encounter of the two anionic reactants. She then extrapolates the same straight line to plan runs at I = 0.50 M, and proposes adding 0.05 M MgSO4 as an independent check of the theory.
1. Check the slope quantitatively. What is the product of charges z_A z_B in the rate-determining step, and is it consistent with the two reactants she has written?
Read the worked answer
Use log k = log k0 + 2A z_A z_B f(I), with A = 0.509 in water at 25 °C and f(I) = √I/(1+√I).
f(I) values: 0.0909, 0.1365, 0.1828, 0.2403.
Slope from first to last point: log(0.093) − log(0.046) = −1.0315 − (−1.3372) = +0.3057, over Δf = 0.1494 → slope = 2.05.
So 1.018 z_A z_B ≈ 2.05 → z_A z_B ≈ +2.
A positive product means the two species forming the encounter complex carry charges of the same sign. That is exactly what S2O8^2− (z = −2) and I− (z = −1) give: (−2)(−1) = +2. Her identification is therefore internally consistent - the salt effect confirms rather than merely assumes the anionic encounter. (Had the reactants been of opposite sign the slope would have been negative, roughly −1 for OH− attack on a monoanion.)
Intercept check: log k0 = −1.3372 − 2.05(0.0909) = −1.524, i.e. k0 ≈ 0.030 M^−1 s^−1 at zero ionic strength - smaller than every measured value, as required for like-charged reactants.
2. Predict k at I = 0.20 M from her line, and evaluate her plan to extrapolate to I = 0.50 M.
Read the worked answer
At I = 0.20: f(I) = 0.4472/1.4472 = 0.3090. log k = −1.524 + 2.05(0.3090) = −0.891 → k ≈ 0.13 M^−1 s^−1.
Judgement on the extrapolations:
- 0.20 M is a mild stretch; the limiting law form √I/(1+√I) is normally quoted as reliable only to I ≈ 0.1 M, so 0.13 is an estimate with perhaps ±10–20 % error, not a prediction.
- 0.50 M is not defensible. Beyond ~0.1 M the Debye–Hückel treatment fails because ion atmosphere thickness becomes comparable to ionic size, specific ion effects (ion pairing of Na+ with S2O8^2−, hydrophobic behaviour of ClO4−) take over, and activity coefficients can pass through a minimum and rise. The linear plot must not be extended through a region where no data exist; the correct plan is to measure k at 0.3–0.5 M.
- Also note that Na+ from the swamping electrolyte is not innocent: ion pairing with persulfate lowers the free S2O8^2− concentration and would depress k at high I, curving the plot downward - a further reason not to extrapolate.
3. She plans a parallel series with 0.05 M MgSO4 instead of NaClO4. Calculate the ionic strength of each additive and predict whether MgSO4 will raise k more. State one reason the prediction could fail.
Read the worked answer
Ionic strengths at 0.05 M:
- MgSO4 (2:2): I = ½(0.05·2² + 0.05·2²) = ½(0.20 + 0.20) = 0.20 M
- Na2SO4 (1:2): I = ½(0.10·1² + 0.05·2²) = ½(0.10 + 0.20) = 0.15 M
Since d log k/dI > 0 here, the divalent salt gives the larger ionic strength and so the larger nominal acceleration: using the fitted line, f(0.20) = 0.309 → log k = −0.891 (k ≈ 0.13) versus f(0.15) = 0.3873/1.3873 = 0.2792 → log k = −0.952 (k ≈ 0.112).
Where the electrostatic argument can fail:
- Mg2+ is a strong ion-pairer toward SO4^2− and S2O8^2−; pairing removes free persulfate and reduces the rate, an effect not captured by I alone. The comparison is only meaningful at matched I with inert 1:1 electrolytes.
- The counter-ion identity enters explicitly at these concentrations; "ionic strength" is a bookkeeping device, not a physical quantity, so matching I does not guarantee matching activity coefficients.