Question

1. The atmospheric concentration of CO2 has exceeded 400 ppm for most of 2016. Water that is in contact with air will absorb CO2 according to Henry’s Law:

[CO2] = kCO2 ∙ PCO2 with kCO2 = 3.1×10–2M/atm at 25°C

a) Use this information to calculate the molarity of CO2(aq) in rain water. Assume that the barometric pressure is 1 atm, the temperature is 25°C, and [CO2] is the equilibrium concentration.

b) Calculate the pH of the rainwater assuming that all absorbed CO2 is in the form of carbonic acid, H2CO3.

You may assume that H2CO3 (aq) is just a monoprotic acid for this problem. Useful information is available in Appendix D on page 1103. (HINT: can you use any approximations or must you solve a quadratic equation? – see Appendix A.3, page 1096)

2. Indicate whether a 1.00 M aqueous solution of each of the following salts is acidic, basic, or neutral at 25 ̊C. Write its chemical name and explain your choice by writing the important chemical equilibrium/equilibria for each salt. (Hint: consider both, cations and anions, as strong or weak acids/bases) Useful information is available in Appendix D on pages 1103‐1104.

a) NaClO2

b) Na2O

c) NH4NO2

d) (NH4)2SO4

e) (NH4)(C2H3O2)

Answer 1

1. The atmospheric concentration of CO2 has exceeded 400 ppm for most of 2016. Water that is in contact with air will absorb CO2 according to Henry’s Law:

[CO2] = kCO2 ∙ PCO2 with kCO2 = 3.1×10–2M/atm at 25°C

a) Use this information to calculate the molarity of CO2(aq) in rain water. Assume that the barometric pressure is 1 atm, the temperature is 25°C, and [CO2] is the equilibrium concentration.

M = H*P

M = 3.1*10^-2 M/atm * (400*10^-6 atm) = 0.0000124 M of CO2

b) Calculate the pH of the rainwater assuming that all absorbed CO2 is in the form of carbonic acid, H2CO3.

0.0000124 of CO2

CO2 in water:

CO2 + H2O --> H2CO3

and

H2CO3 --> H+ HCO3-

Ka = [H+][HCO3-] / [H2CO3]

given Ka = 4.3*10^-7

then[H+] = [HCO3-] = x

[H2CO3] = M-x = 0.0000124 -x

so

Ka = [H+][HCO3-] / [H2CO3]

turns

4.3*10^-7 = x*x/( 0.0000124 -x)

x = 2.1*10^-6

[H+] = x = 2.1*10^-6

pH = -log(2.1*10^-6) = 5.67778