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Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), causing the \(\mathrm{pH}\) 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?

Short Answer

Expert verified
The range of \([\mathrm{H}^+]\) in the raindrops is from \(2.51 \times 10^{-6}\) M to \(6.31 \times 10^{-6}\) M, and the range of \([\mathrm{OH}^-]\) in the raindrops is from \(1.58 \times 10^{-9}\) M to \(3.98 \times 10^{-9}\) M.

Step by step solution

01

Find the range of [H+]

We will use the given pH range (5.2 to 5.6) and the formula for pH (\( \mathrm{pH} = - \log{[\mathrm{H}^+]}\)) to find the range of \([\mathrm{H}^+]\) in the raindrops. We can rewrite the formula as follows to isolate the \([\mathrm{H}^+]\) concentration: \[ [\mathrm{H}^+] = 10^{-\mathrm{pH}} \] Now, we can substitute the given pH values: For pH = 5.2: \[ [\mathrm{H}^+] = 10^{-5.2} \approx 6.31 \times 10^{-6} \,\mathrm{M} \] For pH = 5.6: \[ [\mathrm{H}^+] = 10^{-5.6} \approx 2.51 \times 10^{-6} \,\mathrm{M} \] So, the range of \([\mathrm{H}^+]\) in the raindrops is from \(2.51 \times 10^{-6}\) M to \(6.31 \times 10^{-6}\) M.
02

Calculate the pOH range

We now need to find the range of pOH values corresponding to the given pH range. The relationship between pH and pOH at 25°C is given by the formula \( \mathrm{pH} + \mathrm{pOH} = 14 \). We can calculate the range of pOH values as follows: For pH = 5.2: \[ \mathrm{pOH} = 14 - 5.2 = 8.8 \] For pH = 5.6: \[ \mathrm{pOH} = 14 - 5.6 = 8.4 \] So, the range of pOH in the raindrops is from 8.4 to 8.8.
03

Find the range of [OH-]

We can now use the calculated pOH range (8.4 to 8.8) and the formula for pOH (\( \mathrm{pOH} = - \log{[\mathrm{OH}^-]}\)) to find the range of \([\mathrm{OH}^-]\) in the raindrops. We can rewrite the formula as follows to isolate the \([\mathrm{OH}^-]\) concentration: \[ [\mathrm{OH}^-] = 10^{-\mathrm{pOH}} \] Now, we can substitute the calculated pOH values: For pOH = 8.4: \[ [\mathrm{OH}^-] = 10^{-8.4} \approx 3.98 \times 10^{-9} \,\mathrm{M} \] For pOH = 8.8: \[ [\mathrm{OH}^-] = 10^{-8.8} \approx 1.58 \times 10^{-9} \,\mathrm{M} \] So, the range of \([\mathrm{OH}^-]\) in the raindrops is from \(1.58 \times 10^{-9}\) M to \(3.98 \times 10^{-9}\) M. In conclusion, the range of \([\mathrm{H}^+]\) in the raindrops is from \(2.51 \times 10^{-6}\) M to \(6.31 \times 10^{-6}\) M, and the range of \([\mathrm{OH}^-]\) in the raindrops is from \(1.58 \times 10^{-9}\) M to \(3.98 \times 10^{-9}\) M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH and pOH calculations
Understanding the concepts of pH and pOH is essential in acid-base chemistry. The pH scale, which stands for 'potential of hydrogen', is used to measure the acidity or alkalinity of an aqueous solution. It ranges from 0 to 14, with 7 being neutral. A pH less than 7 indicates acidity while a pH greater than 7 indicates alkalinity.

The pH and pOH scales are interconnected through the equation: \[ \text{pH} + \text{pOH} = 14 \] at 25°C (298 K). This relationship allows us to calculate one if we know the other. Calculating pH from hydrogen ion concentration \( [\text{H}^+] \) involves the negative logarithm base 10: \[ \text{pH} = -\log{[\text{H}^+]} \] Similarly, pOH can be calculated from the hydroxide ion concentration \( [\text{OH}^-] \) using: \[ \text{pOH} = -\log{[\text{OH}^-]} \]

These calculations are crucial for chemists and environmental scientists who need to assess the impact of acid rain, manage water quality, or understand any system where pH plays a significant role.
Carbonic acid in rainwater
Rainwater's pH can provide insight into the atmospheric conditions under which it formed. As carbon dioxide \( CO_2 \) from the air dissolves in rainwater, it forms carbonic acid \( H_2CO_3 \), which can lead to a mild acidity in unpolluted rainwater, typically giving it a pH between 5.2 and 5.6.

The reaction for the formation of carbonic acid is as follows: \[ CO_2(g) + H_2O(l) \leftrightarrow H_2CO_3(aq) \] Even though most \( H_2CO_3 \) dissociates into hydrogen \( H^+ \) and bicarbonate \( HCO_3^- \) ions, the presence of \( H^+ \) is enough to lower the rainwater’s pH from the neutral value of 7. This natural process is exacerbated by pollutants like sulfur dioxide and nitrogen oxides, which can form stronger acids and lead to acid rain with a lower pH, causing environmental harm.
Hydrogen ion concentration
The concentration of hydrogen ions \( [\text{H}^+] \) in a solution determines its acidity. It is expressed in molarity (M), which is the number of moles of hydrogen ions per liter of solution. Using the pH scale, we can identify the \( [\text{H}^+] \) of a substance. For example, a pH of 7 corresponds to a \( [\text{H}^+] \) of \( 10^{-7} \) M.

When we talk about unpolluted rain, the pH ranges from 5.2 to 5.6, which corresponds to a \( [\text{H}^+] \) range from approximately \( 6.31 \times 10^{-6} \) M to \( 2.51 \times 10^{-6} \) M. The precise measurement of \( [\text{H}^+] \) allows us to understand the extent of the rainwater's acidity and, by extension, the potential impact on soil and plant life from acidic precipitation.
Hydroxide ion concentration
In the same way that hydrogen ion concentration informs us about acidity, hydroxide ion concentration \( [\text{OH}^-] \) helps us understand the alkalinity of a solution. The concentration of hydroxide ions is measured in molarity (M), similar to \( [\text{H}^+] \).

For a neutral solution, such as pure water at 25°C, \( [\text{OH}^-] \) and \( [\text{H}^+] \) are equal, each having a concentration of \( 10^{-7} \) M. However, as the presence of carbonic acid in rainwater lowers the pH, the \( [\text{OH}^-] \) is proportionately lower. For the rainwater pH range given (5.2 to 5.6), the corresponding \( [\text{OH}^-] \) ranges from approximately \( 1.58 \times 10^{-9} \) M to \( 3.98 \times 10^{-9} \) M. These values reflect the balance in an aqueous system, as the product of \( [\text{H}^+] \) and \( [\text{OH}^-] \) for water at 25°C is always \( 10^{-14} \) M^2, according to the ion product constant for water.

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Most popular questions from this chapter

Benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) are both derivatives of benzene. Benzoic acid is an acid with \(K_{a}=6.3 \times 10^{-5}\) and aniline is a base with \(K_{a}=4.3 \times 10^{-10}\). (a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\) is a strong electrolyte that dissociates into anilinium ions \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right)\)and chloride ions. Which will be more acidic, a \(0.10 \mathrm{M}\) solution of benzoic acid or a \(0.10 \mathrm{M}\) solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} & \mathrm{NH}_{2}(a q) \rightleftharpoons \\ & \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}+(a q) \end{aligned} $$

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. (i) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the forward reaction is appreciable ( \(K\) much greater than 1 ) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of \(9.95\). Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

Designate the Brønsted-Lowry acid and the Brønsted-Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base of each on the right side. (a) \(\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\) (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons\) \(\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Calculate the pH of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix D): (a) \(0.095 \mathrm{M}\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\), (b) \(0.100 \mathrm{M}\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right)\), (c) \(0.120 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\).

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