A C I D ACID R A I N RAIN

Chemistry Level 3

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Acid rain takes place due to combination of acidic oxides with water and it is an environmental concern all over the world. Assuming rain water is uncontaminated with H N O 3 HNO_3 or H 2 S O 4 H_2SO_4 and is equilibrium with 1.25 × 1 0 − 4 1.25\times10^{-4} a t m atm C O 2 . CO_2. Calculate the p H pH of natural rain water.

Details and assumption:

  • The Henry's law constant ( K h K_h ) = 1.25 × 1 0 6 =1.25\times10^6 t o r r torr

  • ( K a 1 K_{a1} ) of H 2 C O 3 H_2CO_3 = = 4.3 × 1 0 − 7 4.3\times10^{-7}

Give your answer correct to two decimal places


The answer is 5.87.

Harsh Shrivastava by Harsh Shrivastava , 17, India

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1 solution

Harsh Shrivastava
Jun 25, 2018

Relevant wiki: Acids and Bases

By Henry's law :

p C O 2 = x C O 2 p_{CO_2}=x_{CO_2} . K h K_h

x C O 2 = x_{CO_2}= p C O 2 K h = \frac{p_{CO_2}}{K_h}= ( 1.25 × 1 0 − 4 a t m ) × ( 760 t o r r ) 1.25 × 1 0 6 t o r r \frac{(1.25\times10^{-4}atm)\times(760torr)}{1.25\times10^{6}torr}

= 0.76 × 1 0 − 7 = =0.76\times10^{-7}= n C O 2 n t o t a l \frac{n_{CO_2}}{n_{total}}

x H 2 O = ( 1 − 0.76 × 1 0 − 7 ) ≈ 1 = n H 2 O n t o t a l x_{H_2O}=(1-0.76\times10^{-7})\approx1=\frac{n_{H_2O}}{n_{total}}

⇒ n C O 2 n H 2 O ≈ 0.76 × 1 0 − 7 \Rightarrow \frac{n_{CO_2}}{n_{H_2O}}\approx 0.76\times10^{-7}

Molality of C O 2 CO_2 solution

= x C O 2 × 1000 x H 2 O × M w H 2 O = \frac{x_{CO_2}\times1000}{x_{H_2O} \times M_{w_{H_2O}}}

= n C O 2 × 1000 n H 2 O × M w H 2 O = \frac{n_{CO_2}\times1000}{n_{H_2O} \times M_{w_{H_2O}}}

= 0.76 × 1 0 − 7 × 1000 18 = 4.22 × 1 0 − 6 m ≈ 4.22 × 1 0 − 6 M =\frac{0.76\times10^{-7}\times1000}{18}=4.22\times10^{-6}m\approx4.22\times10^{-6}M

Now, dissociation of H 2 C O 3 H_2CO_3

⟶ H 2 C O 3 ⇌ H + + H C O 3 − \longrightarrow H_2CO_3\rightleftharpoons H^++HCO_3^-

and K a K_a = [ H + ] [ H C O 3 − ] [ H 2 C O 3 ] ≈ [ H + ] 2 H 2 C O 3 =\frac{[H^+][HCO_3^-]}{[H_2CO_3]}\approx \frac{[H^+]^2}{H_2CO_3}

∴ [ H + ] = K a [ H 2 C O 3 ] \therefore [H^+]=\sqrt{K_a \, [H_2CO_3]}

= 4.3 × 1 0 − 7 × 4.22 × 1 0 − 6 = 1.345 × 1 0 − 6 M \hspace{1.45cm} = \sqrt{4.3\times10^{-7} \times \, 4.22 \times10^{-6} }\,= 1.345\times10^{-6}M

⇒ pH=5.87 \Rightarrow \fbox{pH=5.87}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I wonder why the complexa tion/formation constant for CuCl was given ;)

Suhas Sheikh - 2 years, 11 months ago

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It was just for confusion for solvers, although I had edited the question.

Harsh Shrivastava - 2 years, 11 months ago

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