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Sagot :
Answer:
pH = 7 ⇒ [H⁺] = 1.0x10⁻⁷ M
pH = 5.6 ⇒ [H⁺] = 2.5x10⁻⁶ M
pH = 3.7 ⇒ [H⁺] = 2.0x10⁻⁴ M
H⁺ concentration in the Hubbard Brook sample is 80 times higher than in unpolluted rainwater.
Explanation:
To answer this problem we need to keep in mind the definition of pH:
- pH = -log[H⁺]
Meaning that after isolating [H⁺] we're left with:
- [H⁺] = [tex]10^{-pH}[/tex]
Now we proceed to calculate [H⁺] for the given pHs:
- pH = 7 ⇒ [H⁺] = [tex]10^{-7}[/tex] = 1.0x10⁻⁷ M
- pH = 5.6 ⇒ [H⁺] = [tex]10^{-5.6}[/tex] = 2.5x10⁻⁶ M
- pH = 3.7 ⇒ [H⁺] = [tex]10^{-3.7}[/tex] = 2.0x10⁻⁴ M
Finally we calculate how many times higher is [H⁺] when pH = 3.7 than when pH = 5.6.
- 2.0x10⁻⁴ / 2.5x10⁻⁶ = 80
Answer:
1. 7 (a neutral solution)
Answer: 10-7= 0.0000001 moles per liter
2. 5.6 (unpolluted rainwater)
Answer: 10-5.6 = 0.0000025 moles per liter
3. 3.7 (first acid rain sample in North America)
Answer: 10-3.7 = 0.00020 moles per liter
The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.
Explanation: -
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