Problem 38
Question
Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6 . What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?
Step-by-Step Solution
Verified Answer
The ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops are approximately \(2.51 \times 10^{-6} \ \mathrm{M}\) to \(6.31 \times 10^{-6} \ \mathrm{M}\) and \(1.59 \times 10^{-9} \ \mathrm{M}\) to \(3.98 \times 10^{-9} \ \mathrm{M}\), respectively.
1Step 1: Recall the definition of pH and relation to concentration of H+ ions
pH is the negative logarithm of the hydrogen ion concentration, i.e., \(\mathrm{pH} = -\log_{10}([\mathrm{H}^+])\). We can rewrite this equation to solve for the concentration of \(\mathrm{H}^{+}\) ions: \([\mathrm{H}^{+}] = 10^{-\mathrm{pH}}\).
2Step 2: Calculate the range of H+ ion concentration
Given the pH range of clean, unpolluted raindrops is between 5.2 and 5.6, we can use the equation from step 1 to find the range of \(\mathrm{H}^{+}\) ion concentration:
Lower bound: \([\mathrm{H}^{+}]_{lower} = 10^{-5.6} \approx 2.51 \times 10^{-6} \ \mathrm{M}\)
Upper bound: \([\mathrm{H}^{+}]_{upper} = 10^{-5.2} \approx 6.31 \times 10^{-6} \ \mathrm{M}\)
So, the range of \(\mathrm{H}^{+}\) ion concentration in the raindrops is approximately \(2.51 \times 10^{-6} \ \mathrm{M}\) to \(6.31 \times 10^{-6} \ \mathrm{M}\).
3Step 3: Use the ion product constant of water to find the OH- ion concentration
The ion product constant of water (\(K_w\)) is given by the equation: \(K_w = [\mathrm{H}^+][\mathrm{OH}^{-}]\), where \(K_w = 1.0 \times 10^{-14}\) at 25°C. We can use this equation to find the range of \(\mathrm{OH}^{-}\) ion concentration:
Lower bound: \([\mathrm{OH}^{-}]_{lower} = \frac{K_w}{[\mathrm{H}^{+}]_{upper}} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6} \ \mathrm{M}} \approx 1.59 \times 10^{-9} \ \mathrm{M}\)
Upper bound: \([\mathrm{OH}^{-}]_{upper} = \frac{K_w}{[\mathrm{H}^{+}]_{lower}} = \frac{1.0 \times 10^{-14}}{2.51 \times 10^{-6} \ \mathrm{M}} \approx 3.98 \times 10^{-9} \ \mathrm{M}\)
So, the range of \(\mathrm{OH}^{-}\) ion concentration in the raindrops is approximately \(1.59 \times 10^{-9} \ \mathrm{M}\) to \(3.98 \times 10^{-9} \ \mathrm{M}\).
In conclusion, the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops are approximately \(2.51 \times 10^{-6} \ \mathrm{M}\) to \(6.31 \times 10^{-6} \ \mathrm{M}\) and \(1.59 \times 10^{-9} \ \mathrm{M}\) to \(3.98 \times 10^{-9} \ \mathrm{M}\), respectively.
Key Concepts
pH CalculationIon ConcentrationChemical Equilibrium
pH Calculation
In the study of acid-base chemistry, understanding and calculating pH is essential. The pH scale measures how acidic or basic a solution is. It ranges from 0, which is very acidic, to 14, which is very basic. A neutral substance, like pure water, has a pH of 7.
To calculate pH, use the formula:
To calculate pH, use the formula:
- \[ \text{pH} = -\log_{10}([\text{H}^+]) \]
- \[ [\text{H}^+] = 10^{-\text{pH}} \]
- For pH 5.6, \([\text{H}^+] = 10^{-5.6} \approx 2.51 \times 10^{-6} \ M\)
- For pH 5.2, \([\text{H}^+] = 10^{-5.2} \approx 6.31 \times 10^{-6} \ M\)
Ion Concentration
In aqueous solutions, ion concentrations are significant to identify solution properties. Specifically, in our atmosphere-related exercise, we're interested in both \([\text{H}^+]\) and \([\text{OH}^-]\) concentrations, which determine whether a solution is acidic or basic.
When you know the hydrogen ion concentration, you can find the hydroxide ion concentration using the ion product constant of water \(K_w\). At 25°C, \(K_w\) is a constant value:
When you know the hydrogen ion concentration, you can find the hydroxide ion concentration using the ion product constant of water \(K_w\). At 25°C, \(K_w\) is a constant value:
- \(K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\)
- \([\text{OH}^-] = \frac{K_w}{[\text{H}^+]}\)
- For \([\text{H}^+] = 6.31 \times 10^{-6}\ M\), \([\text{OH}^-] = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-6}} \approx 1.59 \times 10^{-9} \ M\)
- For \([\text{H}^+] = 2.51 \times 10^{-6}\ M\), \([\text{OH}^-] = \frac{1.0 \times 10^{-14}}{2.51 \times 10^{-6}} \approx 3.98 \times 10^{-9} \ M\)
Chemical Equilibrium
Chemical equilibrium involves a state where the concentrations of reactants and products remain constant over time in a closed system. This concept is crucial to the study of acid-base chemistry, especially when dealing with dissociation processes and reactions in solutions.
In the equilibrium involving water, the self-ionization of water reflects the balance:
In the equilibrium involving water, the self-ionization of water reflects the balance:
- \(2\ H_2O \leftrightarrow H_3O^+ + OH^-\)
- \(K_w = [H_3O^+][OH^-]\) and is always \(1.0 \times 10^{-14}\) at 25°C
- A shift in equilibrium, such as increased \([CO_2]\), can lower the pH, making rain more acidic.
- Maintaining equilibrium ensures that the expressions \([H_3O^+][OH^-] = K_w\) remain constant unless temperature or conditions change.
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