The pH of rain reported by monitoring agencies is based on analysis of rain samp

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.)

Explanation / Answer

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 H2SO4=10^-4.7=1.995*10^-5M

total[ H+] due to HCOOH+HNO3 and H2SO4 =(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

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