Rainwater near a coal-fired power station has a pH of 3.7. Normal rainwater has a pH of 5.5.How many times greater is the concentration of H⁺ ions in the polluted rainwater compared to normal rainwater?

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BlackTom AI · Accepted Answer

To compare how many times greater the hydrogen ion concentration is, remember that pH is defined as pH = -log10[H+]. Therefore, [H+] = 10^(-pH). When you have two pH values, the ratio of their hydrogen ion concentrations is 10^(-pH1) / 10^(-pH2) = 10^(pH2 - pH1).

Here, polluted rain has pH 3.7 and normal rain has pH 5.5. The exponent difference is 5.5 - 3.7 = 1.8, so the ratio is 10^(1.8).

Now evaluate the options in terms of 10^(1.8):
- a. 16 corresponds to 10^(1.204), which would be the ratio if the difference in pH were about 1.204 units. This is not the correct difference for the given pH values.
- b. 63 corresponds to roughly 10^(1.8) since 10^1.8 ≈ 63.1. This matches the calculated ratio using the pH difference 1.8.
- c. 1.8 is the pH difference itself, not the hydrogen ion concentration ratio. The ratio is a power of 10, not the plain difference.
- d. 180 would be close to 10^(2.255), which would require a pH difference of about 2.255. That does not apply here.
- e. 100 equals 10^2, which would correspond to a pH difference of 2.0, not 1.8.

In summary, the key idea is converting pH differences to a ratio of concentrations via 10^(ΔpH). Among the given choices, the one that accurately reflects 10^(1.8) is approximately 63.1. The other options reflect common misunderstandings: treating the difference as a linear factor, or using rounded powers of ten that do not match the actual pH difference given.

Rainwater near a coal-fired power station has a pH of 3.7. Normal rainwater has a pH of 5.5.How many times greater is the concentration of H⁺ ions in the polluted rainwater compared to normal rainwater?

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