Q.ix. The pH of rain water collected in a certain region of Maharashtra on particular day was 5.1. Calculate the $\mathrm{H^+}$ ion concentration of the rain water and its percent dissociation.

Step 1. Find [H+] from pH. $pH=-\log_{10}[H^+]=5.1$, so $\log_{10}[H^+]=-5.1$. Rewriting with a positive mantissa: $-5.1=-6+0.9$, so $[H^+]=10^{-6}\times10^{0.9}$.

Step 2. Evaluate. $10^{0.9}\approx7.943$, so $[H^+]\approx7.94\times10^{-6}$ mol dm-3. This part of the question is fully answerable and is exactly the same style of calculation as this chapter's worked Problem 3.6 and exercise Question 2, item vi.

Step 3. Attempt percent dissociation. Percent dissociation requires $\alpha=[H^+]/c$, where c is the TOTAL (initial, before any dissociation) molar concentration of whatever weak acid is present in the rainwater -- typically taken as dissolved carbonic acid (H2CO3) formed from atmospheric CO2 -- multiplied by 100.

Step 4. Identify the genuine data gap. The problem as printed states only the pH (5.1); it does not state the total carbonic-acid concentration c for this specific rain sample. Without that value, alpha (and hence percent dissociation) cannot be calculated -- and inventing a plausible-sounding concentration would risk presenting a fabricated number as if it were given data, which this platform's sourcing standards do not allow. …

Compute [H+] directly from the given pH (a complete, self-contained calculation); for percent dissociation, identify explicitly that the required total-concentration input is mi …

• Assuming a 'typical' or 'normal' rainwater carbonic-acid concentration (e.g. from unrelated general chemistry knowledge) to force a percent-dissociation number -- this problem's own printed text does not supply that figure, so doing so would present an invented number as if it were given data. …