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Rainwater is acidic because \(\mathrm{CO}_{2}(\mathrm{~g})\) dissolves in the water, creating carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{5}\). If the rainwater is toe acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, \(\mathrm{CaCO}_{3}\) ). Calculate the concentrations of carbonic acid, bicarbonate ion \(\left(\mathrm{HCO}_{3}{ }^{-}\right)\)and carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) that are in a raindrop that has a \(\mathrm{pH}\) of \(5.60\), assuming that the sum of all three species in the raindrop is \(1.0 \times 10^{-5} \mathrm{M} .\)

Short Answer

Expert verified
In a raindrop with a pH of 5.60, the concentrations of carbonic acid, bicarbonate ion, and carbonate ion can be determined by considering the dissociation equations and provided information. Using the pH, we calculate the hydrogen ion concentration \(\mathrm{[H^+]} = 10^{-\mathrm{pH}} \). The dissociation constants here are, \(K_{a_1}\) and \(K_{a_2}\). Set up and solve the system of three equations using these dissociation constants and the sum of the concentrations of each species \(1.0 \times 10^{-5} \mathrm{M}\). By solving this system of equations for x, y, and z, we get the desired concentrations of carbonic acid (\(\mathrm{[H_2CO_3]}\)), bicarbonate ion (\(\mathrm{[HCO_3^-]}\)), and carbonate ion (\(\mathrm{[CO_3^{2-}]}\)).

Step by step solution

01

Calculate the hydrogen ion concentration from the pH

The pH is given as 5.60, so we can use this to find the hydrogen ion concentration (\(\mathrm{[H^+]}\)) using the formula: \[ \mathrm{pH} = -\log (\mathrm{[H^+]}) \] Solving for \(\mathrm{[H^+]}\) we get: \[ \mathrm{[H^+]} = 10^{-\mathrm{pH}} \]
02

Calculate the concentrations using the dissociation constants

The dissociation constants for the two reactions are: - First dissociation constant: \(K_{a_1} = \frac{\mathrm{[HCO_3^-][H^+]}}{\mathrm{[H_2CO_3]}}\) - Second dissociation constant: \(K_{a_2} = \frac{\mathrm{[CO_3^{2-}][H^+]}}{\mathrm{[HCO_3^-]}}\) We know the sum of the concentrations of all three species is given by \(1.0 \times 10^{-5} \mathrm{M}\), so: \(\mathrm{[H_2CO_3]} + \mathrm{[HCO_3^-]} + \mathrm{[CO_3^{2-}]} = 1.0 \times 10^{-5}\) Using these dissociation equations and the provided information, we can solve for the concentrations of each species.
03

Solve for the individual concentrations

Let \(\mathrm{x = [H_2CO_3]}\), \(\mathrm{y = [HCO_3^-]}\), and \(\mathrm{z = [CO_3^{2-}]}\). With the known dissociation constants and the sum of the concentrations, we have a system of three equations with three unknowns: \[ K_{a_1} = \frac{y [H^+]}{x} \] \[ K_{a_2} = \frac{z [H^+]}{y} \] \[ x + y + z = 1.0 \times 10^{-5} \] Solve this system of equations for x, y, and z.
04

Report the calculated concentrations

Once you have found the values of x, y, and z, we have the concentrations of each species: - Carbonic acid concentration: \(\mathrm{[H_2CO_3] = x}\) - Bicarbonate ion concentration: \(\mathrm{[HCO_3^-] = y}\) - Carbonate ion concentration: \(\mathrm{[CO_3^{2-}] = z}\) These concentrations should sum up to \(1.0 \times 10^{-5} \mathrm{M}\) as per the information given in the exercise.

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Key Concepts

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

pH Calculation
Understanding how to calculate pH is crucial when studying acid rain chemistry, as it directly correlates to the acidity or basicity of a solution. The pH is a scale that ranges from 0 to 14, with lower values indicating higher acidity. To find the hydrogen ion concentration \( \mathrm{[H^+]} \), which is essential in the process of calculating pH, the following mathematical relationship is used:

\[ \mathrm{pH} = -\log(\mathrm{[H^+]}) \]

This formula indicates that pH is the negative logarithm of the hydrogen ion concentration. For a raindrop with a pH of 5.60, the calculation would be:

\[ \mathrm{[H^+]} = 10^{-5.60} \]

This result shows the concentration of hydrogen ions present in the raindrop, giving us a foundational value to further assess the raindrop's chemistry.
Dissociation Constants
The dissociation constants, \( K_{a} \), are vital parameters in acid rain chemistry that quantify the extent to which a compound dissociates in water, forming ions. These constants are unique for each acid and its corresponding stages of ionization. In our example, we look at carbonic acid, \( \mathrm{H}_2\mathrm{CO}_3 \), which can dissociate into bicarbonate ions, \( \mathrm{HCO}_3^- \), and carbonate ions, \( \mathrm{CO}_3^{2-} \).

The first dissociation constant, \( K_{a_1} \), concerns the conversion of carbonic acid to bicarbonate ion, while the second, \( K_{a_2} \), pertains to the conversion of bicarbonate ion to carbonate ion. Their expressions are given by:
\[ K_{a_1} = \frac{\mathrm{[HCO_3^-][H^+]}}{\mathrm{[H_2CO_3]}} \]
\[ K_{a_2} = \frac{\mathrm{[CO_3^{2-}][H^+]}}{\mathrm{[HCO_3^-]}} \]

These constants allow us to calculate the balance of different species of dissolved carbonates in the water, which is essential for predicting the behavior of acid rain when it interacts with substances like limestone and seashells.
Carbonate System Equilibrium
When discussing the equilibrium of the carbonate system in the context of acid rain, we analyze the dynamic balance between carbonic acid, bicarbonate ion, and carbonate ion. In essence, the system is governed by the concentrations of these species, which are interconnected by the dissociation reactions. The total concentration is the sum of all three, which can be represented as:

\[ \mathrm{[H_2CO_3]} + \mathrm{[HCO_3^-]} + \mathrm{[CO_3^{2-}]} = \text{total concentration} \]

By setting up a system of equations using the dissociation constants and the total concentration specified, we can solve for the individual concentrations in the carbonate system. This allows us to understand the buffering capacity of the raindrop against changes in pH and how it reacts when coming into contact with minerals.
Carbonic Acid Concentration
The concentration of carbonic acid in a solution, like our raindrop example, is directly tied to the amount of dissolved carbon dioxide. In the presence of water, carbon dioxide forms carbonic acid, which is a weak acid that partially dissociates into bicarbonate and carbonate ions. The balance between these species is essential in environmental chemistry, especially when considering the effects of acid rain on ecosystems and man-made structures.

By applying the previously described formulas and knowing the total concentration as well as the dissociation constants, we can determine the concentration of carbonic acid in the raindrop. This concentration is a starting point to predict how acid rain might contribute to the erosion of calcium carbonate-based materials, such as limestone and seashells, which are components of many natural and historical artifacts.

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Most popular questions from this chapter

(a) Calculate the percent ionization of \(0.125 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right)\). (b) Calculate the percent ionization of \(0.125 \mathrm{M}\) lactic acid in a solution containing \(0.0075 \mathrm{M}\) sodium lactate. Buffers (Section 17.2)

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. (b) The pH at the beginning of the titration is lower for the weak acid than the strong acid. (c) The \(\mathrm{pH}\) at the equivalence point is 7 no matter which acid is titrated.

Furoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of (a) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\). of \(0.22 \mathrm{M} \mathrm{NaC} \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL}\). (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500 \mathrm{~L}\) of \(0.0850 \mathrm{M} \mathrm{HC} \mathrm{H}_{3} \mathrm{O}_{3}\).

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\left(K_{u p}=4 \times 10^{-10}\right)\) in 1.0 L of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) because the solution will be quite basic.)

From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2}(a q)\) and \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) that are present at equilibrium after dissolving 1.25 \(\mathrm{g} \mathrm{NiCl}_{2}\) in 100.0 \(\mathrm{mL}\) of 0.20 \(\mathrm{MN} \mathrm{H}_{3}(a q) .\)

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