Question

The pH of rain reported by monitoring agencies is based on analysis of rain samples collected weekly in buckets. The weekly collection schedule is fine for HNO₃ and H₂SO₄, which do not degrade; however formic acid (HCOOH) is rapidly consumed by bacteria in the buckets and is therefore entirely removed before analysis. The Henry’s law and acid dissociation equilibria for HCOOH are:

HCOOH_{(g) <-- -->} HCOOH_(aq)               H = 3.7x10³ M atm

HCOOH_{(aq) <-- -->}  HCOO^- + H⁺ K₁ = 1.8x10^-4 M

If the monitoring agency reports a rainwater pH of 4.7, calculate the true pH of the rain. Assume 1 ppbv HCOOH in the atmosphere, a typical value for the eastern U.S. (Hint: start by assuming charge balance.)

Answer 1

According to Henry's law ,at a constant temperature, the amount of gas dissolved in a solvent is directly proportional to the p(gas)(partial pressure) of the gas .

So, H= henry constant=concentration of gas in aqueous phase/partial pressure of gas

H=[HCOOH]aq/p(COOH)

p(HCOOH)*H=[HCOOH]aq

C(HCOOH)=1ppbV=1 micro g/L=10^-6 g/L /(46.025g/mol) [molar mass of formic acid=46.025 g/mol]

C(COOH)atm=0.0217*10^-6 mol/L or M

p(HCOOH)=cRT [ideal gas law]

p(HCOOH)=(0.0217*10^-6 mol/L )*0.0821L atm/K.mol*298K=0.531*10^-6 atm

So,p(HCOOH)*H=[HCOOH]aq=0.531*10^-6 atm*(3.7*10^3 Matm)=1.964*10^-3 M

HCOOH_{(aq) <---->} HCOO^- + H⁺

Ka=1.8*10^-4 M=[HCOOH]aq/[H+][HCOO-]

At equilibrium,

[HCOOH]aq=1.964*10^-3 M-x

[H+]=[HCOO-]=x

1.8*10^-4 M=[[H+][HCOO-]/[HCOOH]aq=x^2/1.964*10^-3 M-x

Ignoring x with respect to 4.166M as x<<<4.166 ,very less dissociation for weak acid

1.8*10^-4 M=x^2/1.964*10^-3 M

x=5.946*10^-4M=[H+]

Also ,pH=-log[H+]

[H+] due to HNO₃ and H₂SO_{4=10^-4.7=1.995*10^-5M}

_{total[ H+] due to HCOOH+}HNO₃ and H₂SO_{4 =(}1.995*10^-5M)+ 5.946*10^-4M=6.145*10^-4M(approx)

pH=-log 6.145*10^-4M=3.211

true pH=3.2