Part 1Raindrops found in an unpolluted atmosphere (where CO2 (8) concentrations are0.000390 atm) have a pH of 5.62. What is value of pOH?Charge balance: [H*] = [HCO3] + 2[CO32] + [OH']Helpful reactions:= 4.5 × 10:7= 4.7 x 10-111.0 × 10-14102 M/atmH₂CO3 + H₂O H3O+ + HCO3²-HCO3 + H₂0 H3O+ + CO3²-H₂O H₂O + OHCO₂ (8) = H₂CO3Ka1Ka2KwKH = 3.38 xPHH

The question is based on the concept of the pH of the solution.It is defined as a negative logarithm…

Part 1 Raindrops found in an unpolluted atmosphere (where CO₂ (g) concentrations are 0.000390 atm) have a pH of 5.62. What is value of pOH? Charge balance: [H*] = [HCO3] + 2[CO3²] + [OH] Helpful reactions: = 4.5 × 10¹7 X = 4.7 x 10-11 = 1.0 × 10-14 102 M/atm H₂CO3 + H₂O H3O+ + HCO3²- HCO3 + H₂O = H³O+ + CO3²- H₂O H3O+ OH₁ CO₂ (g) = H₂CO3 Kal Ka2 Kw KH = 3.38 x F H H

Part 1 Raindrops found in an unpolluted atmosphere (where CO₂ (g) concentrations are 0.000390 atm) have a pH of 5.62. What is value of pOH? Charge balance: [H*] = [HCO3] + 2[CO3²] + [OH] Helpful reactions: = 4.5 × 10¹7 X = 4.7 x 10-11 = 1.0 × 10-14 102 M/atm H₂CO3 + H₂O H3O+ + HCO3²- HCO3 + H₂O = H³O+ + CO3²- H₂O H3O+ OH₁ CO₂ (g) = H₂CO3 Kal Ka2 Kw KH = 3.38 x F H H

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The Keq for the reaction of water with water is called Kw H₂O + H₂O → H30+1 + OH-1 [H3O+¹][OH-¹] Kw [H₂O]² The inverse logarithm, or -log, of [H3O+¹][OH-¹] = 10-14 is pH + pOH = 14 1. What happens to the [H3O+¹] when acid is added to neutral water? the [H3O+¹] goes [H3O+¹][OH-¹] = (1 x 10-7)² = 1 x 10-14 2. What happens to the pH when acid is added to neutral water? an example of an increased [H3O+¹] is [H3O+¹] = 1 x 106. Thus pH = -log[106] = 6 thus, adding acid makes the pH go 3. What happens to the [OH-¹] when acid is added to neutral water? in the equation [H3O+¹][OH^¹] = 10-¹4, when [H3O+¹] increases, the [OH-¹] goes 4. What happens to the pOH when acid is added to neutral water? when [OH-¹] decreases, the pOH goes this can also be seen using pH + pOH = 14 5. What happens to the Kw when acid is added to neutral water ? because it's a constant. a. down b. up c. no change since the pOH = -log[OH-¹]