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. Assuming levels of atmospheric CO2 are 390 ppm CO2, what is the pH of rainwa...

Mar 19, 2024

. Assuming levels of atmospheric CO2 are 390 ppm CO2, what is the pH of rainwater due to the presence of carbon dioxide? (2 points) Some estimates are for atmospheric carbon dioxide levels to double in the future. What would be the pH of rainwater if this happens? (0.5 points)

Solution

Identify the reaction: The dissolution of CO2 in water forms carbonic acid (H2CO3), which is a weak acid. The relevant chemical reactions are: CO2(g) + H2O(l) ⇌ H2CO3(aq) ⇌ HCO3^-(aq) + H^+(aq) ⇌ CO3^2-(aq) + 2H^+(aq)

Determine the equilibrium constant: The first dissociation of carbonic acid (H2CO3 to HCO3^- and H^+) has an equilibrium constant (Ka1) of about 4.3 × 10^-7 at 25°C

Calculate the hydrogen ion concentration: Assuming that the concentration of CO2 in rainwater is proportional to its partial pressure, we can use Henry's law to find the concentration of dissolved CO2. The Henry's law constant for CO2 at 25°C is 3.3 × 10^-2 mol/L·atm. At 390 ppm CO2, the partial pressure of CO2 is 390 × 10^-6 atm. Thus, [CO2] = 3.3 × 10^-2 mol/L·atm × 390 × 10^-6 atm = 1.287 × 10^-5 mol/L. Using the equilibrium constant, we can find [H^+] = √(Ka1 × [CO2]) = √(4.3 × 10^-7 × 1.287 × 10^-5) ≈ 2.24 × 10^-6 M

Calculate the pH: pH = -log[H^+] = -log(2.24 × 10^-6) ≈ 5.65

Answer

pH of rainwater with 390 ppm CO2 is approximately 5.65.

Key Concept

The pH of rainwater is influenced by the dissolution of CO2 forming carbonic acid, which dissociates to release H^+ ions.

Explanation

The pH is calculated using the concentration of hydrogen ions, which is found by applying Henry's law to determine the concentration of dissolved CO2 and then using the equilibrium constant for the dissociation of carbonic acid.

Solution

Predict the effect of doubled CO2 levels: If atmospheric CO2 levels double, the partial pressure of CO2 would be 780 ppm

Calculate the new CO2 concentration: Using Henry's law again, [CO2] = 3.3 × 10^-2 mol/L·atm × 780 × 10^-6 atm = 2.574 × 10^-5 mol/L

Calculate the new hydrogen ion concentration: [H^+] = √(Ka1 × [CO2]) = √(4.3 × 10^-7 × 2.574 × 10^-5) ≈ 3.17 × 10^-6 M

Calculate the new pH: pH = -log[H^+] = -log(3.17 × 10^-6) ≈ 5.50

Answer

pH of rainwater with doubled CO2 levels (780 ppm) is approximately 5.50.

Key Concept

Doubling atmospheric CO2 levels increases the concentration of H^+ ions in rainwater, resulting in a lower pH.

Explanation

The pH decrease is due to the higher concentration of dissolved CO2, which leads to more carbonic acid formation and dissociation, increasing the concentration of hydrogen ions.

There is two liters of 0.50 mol·L-1 NH3·H2O and 2 liter of 0.50 mol·L-1 HCl in a laboratory. A technician wants to use them to prepare a buffer of pH=9.00 without the addition of water. How many liters of buffer can the technician prepare at most? What are the concentrations of NH3·H2O and NH4 + in the buffer? (pKb NH3·H2O =4.74.)

Solution

Calculate the moles of NH3·H2O and HCl: Before preparing the buffer, we need to know the amount of NH3·H2O and HCl available

Moles of NH3·H2O = Volume (L) × Concentration (mol/L) = 2 L × 0.50 mol/L = 1 mol. Similarly, Moles of HCl = 2 L × 0.50 mol/L = 1 mol.

Reaction between NH3·H2O and HCl: NH3·H2O will react with HCl to form NH4+ and Cl-. The reaction is NH3·H2O + HCl → NH4+ + Cl- + H2O

Since both NH3·H2O and HCl are present in equal moles, they will completely react with each other to form 1 mol of NH4+.

Use the Henderson-Hasselbalch equation: The pH of the buffer is related to the pKa of the conjugate acid (NH4+) and the ratio of the concentrations of the conjugate base (NH3·H2O) and conjugate acid (NH4+)

pH = pKa + log([NH3·H2O]/[NH4+]), where pKa = 14 - pKb = 14 - 4.74 = 9.26.

Calculate the ratio of NH3·H2O to NH4+: Rearrange the Henderson-Hasselbalch equation to find the ratio of NH3·H2O to NH4+

[NH3·H2O]/[NH4+] = 10^(pH - pKa) = 10^(9.00 - 9.26) = 10^(-0.26) ≈ 0.55.

Determine the buffer capacity: Since the ratio of NH3·H2O to NH4+ is 0.55, for every 1.55 moles of NH4+, we need 0.55 moles of NH3·H2O

We have 1 mole of NH4+ from the reaction, so we can use 0.55 moles of NH3·H2O to create the buffer.

Calculate the volume of NH3·H2O required: To find out how much NH3·H2O is needed to achieve the desired ratio, we use the moles and concentration

Volume of NH3·H2O = Moles / Concentration = 0.55 mol / 0.50 mol/L = 1.1 L.

Calculate the final concentrations in the buffer: The total volume of the buffer is the sum of the volumes of NH3·H2O and HCl used

Total buffer volume = 1.1 L (NH3·H2O) + 2 L (HCl) = 3.1 L.

Concentration of NH4+: Since all the HCl reacts, the concentration of NH4+ is based on the initial moles of HCl

[NH4+] = Moles of NH4+ / Total buffer volume = 1 mol / 3.1 L ≈ 0.32 mol/L.

Concentration of NH3·H2O: The concentration of NH3·H2O is based on the remaining moles after the reaction

[NH3·H2O] = (Initial moles of NH3·H2O - Moles used) / Total buffer volume = (1 mol - 0.55 mol) / 3.1 L ≈ 0.15 mol/L.

Answer

The technician can prepare at most 3.1 liters of buffer. The concentrations of NH3·H2O and NH4+ in the buffer are approximately 0.15 mol/L and 0.32 mol/L, respectively.

Key Concept

The preparation of a buffer solution involves a reaction between a weak base and its conjugate acid, and the use of the Henderson-Hasselbalch equation to determine the required concentrations.

Explanation

The buffer capacity is determined by the stoichiometry of the reaction between NH3·H2O and HCl, and the desired pH is achieved by adjusting the ratio of NH3·H2O to NH4+ using the Henderson-Hasselbalch equation. The final concentrations are calculated based on the total volume of the buffer and the moles of each component.

In a water sample with pH=6.0, [SO4 2- ] =1.0*10-3 mol·L-1 and you can smell H2S. Assuming that PH2S=1.0*10-6 atm, please calculate the Eh and pE for this system (25 oC). The equation is: SO4 2- + 10H + + 8e → H2S + 4H2O and lgK=42.

Solution

Write the Nernst equation for the reaction: The Nernst equation relates the reduction potential of a chemical reaction to the standard electrode potential, temperature, and activities (or concentrations) of the reactants and products. For the given reaction, the Nernst equation is:

E = E^0 - \frac{0.05916}{n} \log \frac{[\text{H}^+]^{10}}{[\text{SO}_4^{2-}][\text{H}_2S]}

where

n

is the number of electrons transferred in the reaction, which is 8

Calculate the standard electrode potential \( E^0 \): The standard electrode potential can be calculated using the equation

\log K = \frac{nE^0}{0.05916}

, where

K

is the equilibrium constant of the reaction. Given

\log K = 42

, we can solve for

E^0

E^0 = \frac{42 \times 0.05916}{8} = 0.3695 \text{ V}

Calculate the activity of \( \text{H}^+ \) from pH: The activity of

\text{H}^+

is given by

10^{-\text{pH}}

. With a pH of 6.0, the activity of

\text{H}^+

10^{-6} \text{ mol/L}

Calculate the partial pressure of \( \text{H}_2S \): The partial pressure of

\text{H}_2S

is given as

1.0 \times 10^{-6} \text{ atm}

. To find the concentration of

\text{H}_2S

in mol/L, we use Henry's law:

[\text{H}_2S] = \text{PH}_2S / K_H

. Assuming

K_H

is approximately

1 \times 10^{-3} \text{ M/atm}

at 25°C, we get

[\text{H}_2S] = 1.0 \times 10^{-3} \text{ M}

Calculate the electrode potential \( E \): Using the Nernst equation from step 1 with the values from steps 2, 3, and 4:

E = 0.3695 \text{ V} - \frac{0.05916}{8} \log \frac{(10^{-6})^{10}}{(1.0 \times 10^{-3})(1.0 \times 10^{-3})}

After calculating, we find that

E \approx 0.14 \text{ V}

Calculate the pE: The pE is calculated using the equation

pE = -\log [e^-]

, where

[e^-]

is the concentration of electrons. The pE is related to the electrode potential by the equation

pE = \frac{E}{0.05916}

. Using the value of

E

from step 5:

pE = \frac{0.14}{0.05916}

After calculating, we find that

pE \approx 2.36

Answer

Eh = 0.14 V, pE = 2.36

Key Concept

The Nernst equation is used to calculate the electrode potential (Eh) and pE for a redox system in equilibrium.

Explanation

The Nernst equation takes into account the standard electrode potential, temperature, and the activities of the reactants and products to determine the actual electrode potential and pE of the system.

A graduate student made some mistakes in water sampling because he lacks experience. After the onsite measurement showed that temperature is 25 ℃ and pH=7.8, the water sample bottle was put on a cart and brought back to the lab exposed to sunshine. The pH went up to 10.2 and in the headspace of the bottle, PO2 was found to be 0.4 atm. The temperature is still 25 ℃. What caused the increase in pH and Po2? What’s the difference between the pE measured in lab and onsite? (We know that for equation 1/4O2+H++e → 1/2H2O, pE0 =20.8)

Solution

Identify the changes in conditions: The pH of the water sample increased from 7.8 to 10.2 after being exposed to sunshine, and the partial pressure of oxygen (PO2) in the headspace of the bottle increased to 0.4 atm

Explain the pH increase: The increase in pH could be due to photosynthetic activity by algae or other microorganisms present in the water sample. Photosynthesis consumes CO2, which decreases the concentration of carbonic acid and increases pH

Explain the PO2 increase: The increase in PO2 could be a result of oxygen production during photosynthesis. As photosynthetic organisms produce oxygen, the concentration of oxygen in the headspace of the bottle would increase

Discuss the pE difference: The pE onsite would be lower than the pE measured in the lab due to the onsite conditions being closer to equilibrium with atmospheric oxygen levels. The lab conditions, with higher PO2, would lead to a higher pE value

Answer

The increase in pH and PO2 is likely due to photosynthetic activity. The pE measured in the lab would be higher than the pE measured onsite due to the increased PO2.

Key Concept

Photosynthesis affects pH and PO2 in water samples

Explanation

Photosynthesis consumes CO2, which decreases the concentration of carbonic acid and increases pH, and produces O2, which increases PO2. The pE value is influenced by the partial pressure of oxygen.