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Scilla [17]
3 years ago
7

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

um equation.
CO2(g) <----> CO2(aq) The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. K=0.032 M/atm
For a CO2 partial pressure of 4.4×10-4 atm in the atmosphere, what is the pH of water in equilibrium with the atmosphere?
Note: Since carbonic acid is primarily dissolved CO2, the concentration of H2CO3 can be taken as equal to that of dissolved CO2.
My online homework is due tonight and I am hopelessly lost on how to solve this problem. :(
Chemistry
1 answer:
Dvinal [7]3 years ago
6 0
C = pK 
<span>C = 4.4E-4*0.032 = about 1.41E-5 </span>

<span>H2CO3 ==> H^+ + HCO3^- </span>

<span>k1 = (H^+)(HCO3^-)/(H2CO3) </span>
<span>(H^+)= (HCO3^-) = x </span>
<span>(H2CO3) = 1.41E-5 </span>
<span>Solve for x = (H^+) and convert to pH. </span>
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=C_3H_4O_3

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When percentage composition is given, and asked for the empirical formula, it is simplest to  assume 100 g of material. Thus,

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Mass H = 4.58 g.  Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g.  Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

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Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

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