pH of a raindrop as a function of time

Q1 from the class notes, fully solved: 5 ppb SO₂ being oxidised by H₂O₂ and O₃ inside a cloud drop, with HNO₃ and NH₃ around. Below: the method step by step, the real numbers at every minute, a simulator you can play with, and a ready-made Excel workbook that builds the pH profile with live formulas.

⬇ Download the Excel workbook (raindrop-pH-profile.xlsx)

Reading the question

Q1. S(IV) in the presence of H₂O₂ (and O₃). Calculate the pH of the raindrop as a function of time.

What we have to assume (state these in your answer)

1. Closed cloud parcel. The air and the drops inside it exchange gases with each other but not with outside air. So the 1 ppb of H₂O₂ is all the H₂O₂ there is, and it can run out.
2. Liquid water content $L = 0.2\ \text{g/m}^3$ (a typical cloud). This is the one number the question does not give. It is the number that makes the initial pH come out at 6.08, which matches the "assume pH = 6 at t = 0" in the notes. Try other values in the simulator.
3. T = 298 K, with Henry's law and acid–base equilibria established instantly. Chemistry is slow (minutes) compared with dissolving (seconds).
4. Rate laws and constants from Seinfeld & Pandis, Ch. 7, which is the standard source for this problem. If your professor used different constants, change the yellow cells in the Excel file.
5. CO₂ and HCO₃⁻ are ignored, as in the charge balance in the notes. At pH 5–6 they add under 6 µM, compared with ~300 µM for the other ions.

Step 1: ppb in air → mol per litre of drop water

Everything in the drop is in mol/L (M), but the question gives gas mixing ratios. The bridge between them is how much water there is per volume of air.

Most gases do not dissolve completely. Henry's law shares each one between gas and drop:

where $H^*$ is the effective Henry's constant. It includes the ions the gas forms in water, so it depends on pH:

Step 2: How the dissolved S(IV) splits up

Inside the drop, S(IV) exists as three forms that convert between each other instantly:

Combining the two equilibria with $[\text{S(IV)}]_{aq}=[\text{SO}_2\cdot\text{H}_2\text{O}]+[\text{HSO}_3^-]+[\text{SO}_3^{2-}]$, and writing $D=[\text{H}^+]^2+K_{s1}[\text{H}^+]+K_{s1}K_{s2}$, gives:

These are the formulas in your notes, except that every one of them is multiplied by $[\text{S(IV)}]_{aq}$ (the notes forgot it on SO₃²⁻), and $[\text{S(IV)}]_{aq}$ is the dissolved part from step 1, not the 5 ppb total.

Step 3: The two oxidation rates

Your notes write $r = -\dfrac{d\,\text{S(IV)}}{dt} = \dfrac{d\,\text{S(VI)}}{dt} = r_{O_3}+r_{H_2O_2}$. To get numbers you need the rate laws, which are the standard ones (Seinfeld & Pandis, 298 K):

The two lines cross at about pH 5.95. Our drop starts at 6.08, just above the crossing, so ozone is ahead for the first ~2 minutes. Once the pH falls below 5.95, H₂O₂ is ahead for the rest of the run. This crossover is the whole story of the curve's shape. Between pH 6 and 5 the ozone rate falls 90×, while the H₂O₂ rate changes by under 10%.

Step 4: Stepping through time (Euler method)

You can't integrate $dS/dt$ on paper because the rate depends on pH, which depends on S. So we march forward in small steps of $\Delta t$ (1 min = 60 s), assuming the rate stays constant within each step:

These are the totals (gas + drop, in µM of water). HNO₃ and NH₃ totals don't change; only how much of the NH₃ sits in the drop changes, and that depends on pH.

Step 5: Charge balance → the new pH

A drop is electrically neutral, so positive charge equals negative charge. This is the equation in your notes:

Every term is a function of $[\text{H}^+]$, given the totals at that moment:

Move everything to one side: $F(\text{pH}) = \text{cations} - \text{anions}$. The pH we want is where $F=0$. You can't rearrange this for pH by hand, so we search for it:

t = 0 and t = 1 min, by hand

The full pH profile (0 → 50 min)

Play with it

This is the same model, running live in your browser. Change one thing at a time and predict what happens before you look.

Doing it in MS Excel (the pH profile)

⬇ raindrop-pH-profile.xlsx already contains everything below with live formulas. Open it, see the chart, and change any yellow input. Read on to understand it, or to build it yourself from scratch.

Step 1: The Inputs sheet and named cells

Put each constant in a cell, then name it: click the cell, type the name in the Name Box (left of the formula bar) and press Enter. Named cells make formulas readable, e.g. =Ka_S1*G2/J2 instead of =Inputs!$B$27*G2/J2.

Step 2: The Profile columns

Row 2 is t = 0. Every cell below is in mol/L.

Fill row 3 across, then select row 3 and drag the fill handle down to row 52 (t = 50).

Step 3: Getting the pH in each row. Pick one method.

Sub SolveAllRows()
    Dim r As Long
    With Worksheets("GoalSeek_Method")
        For r = 5 To 55
            .Cells(r, "F").Value = 6
            .Cells(r, "S").GoalSeek Goal:=0, ChangingCell:=.Cells(r, "F")
        Next r
    End With
End Sub

Step 4: The chart

1. Select A1:A52, then hold Ctrl and select F1:F52.
2. Insert → Scatter → Scatter with Smooth Lines and Markers. Don't use a "Line" chart, because it treats time as text labels.
3. Axis titles: Time (min) and pH. Set the y-axis to 4.8–6.2 so the shape is visible.
4. Title: pH of raindrop vs time: S(IV) oxidation by H₂O₂ and O₃.

How to write it in the exam

1. Assumptions: closed cloud parcel, LWC = 0.2 g/m³, T = 298 K, Henry's law and acid–base equilibria reached instantly, CO₂ neglected.
2. Conversion: $c_{tot}=\xi\times10^{-9}/(RTL)$; 1 ppb = 204.5 µM. Share between gas and drop with $H^*RTL/(1+H^*RTL)$.
3. Speciation: $[\text{HSO}_3^-]=\dfrac{K_{s1}[\text{H}^+]}{[\text{H}^+]^2+K_{s1}[\text{H}^+]+K_{s1}K_{s2}}[\text{S(IV)}]_{aq}$ (and the other two forms).
4. Rates: $r_{H_2O_2}=k[\text{H}^+][\text{H}_2\text{O}_2][\text{HSO}_3^-]/(1+K[\text{H}^+])$; $r_{O_3}=(k_0[\text{SO}_2\cdot\text{H}_2\text{O}]+k_1[\text{HSO}_3^-]+k_2[\text{SO}_3^{2-}])[\text{O}_3]$.
5. Mass balance (Euler, Δt = 1 min): $\text{S(IV)}_{i+1}=\text{S(IV)}_i-r\Delta t$, $\text{S(VI)}_{i+1}=\text{S(VI)}_i+r\Delta t$, H₂O₂ and O₃ consumed likewise.
6. Charge balance solved for [H⁺] at each step (Goal Seek or bisection).
7. Result: pH 6.08 (0) → 5.67 (5) → 5.49 (10) → 5.28 (20) → 5.17 (30) → 5.11 (40) → 5.07 (50 min). O₃ dominates for the first ~2 min, then H₂O₂ dominates until it is used up. NH₃ uptake buffers the drop. Draw the curve: a steep fall, then a flattening near 5.1.

Quick quiz

Fixes to the handwritten notes

1. "5 ppb → 5 µg/L": wrong for this problem. These are gas-phase mixing ratios. Convert with Henry's law and $RTL$ (section 2).
2. Page 2: "dS(IV)/dt = r_O₃ + r_H₂O₂" is missing a minus sign: $\dfrac{d\,\text{S(IV)}}{dt}=-(r_{O_3}+r_{H_2O_2})$. Page 1 has it right.
3. [SO₃²⁻] formula: the numerator should be $K_{a1}K_{a2}\,[\text{S(IV)}]_{aq}$. The S(IV) term is missing.
4. [S(IV)] in the speciation formulas is the dissolved amount, not the 5 ppb total. You have to do the gas/drop split first.
5. "Assuming pH = 6 at t = 0": you don't have to assume it. The charge balance at t = 0 gives 6.08 (with LWC 0.2 g/m³), so state that as a check.
6. [NH₄⁺] is not constant. It has to be recalculated at every step, because more NH₃ dissolves as the drop acidifies. This is what makes the curve flatten.
7. "H₂SO₃(aq)" is really SO₂·H₂O. A H₂SO₃ molecule has never been found in solution. Writing it as a label is fine.

Constants: Seinfeld & Pandis, Atmospheric Chemistry and Physics, Ch. 7 (Tables 7.2–7.4 and the S(IV) oxidation rate expressions). Every number on this page was produced by the same model as the Excel file, and the two were checked against each other.