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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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Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y
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The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = 10^{-3.5} = 3.162 * 10^{-4}

From the Ice table

3.162 * 10^{-4} = \frac{x + H^+}{[aspirin]}  = \frac{x^{2} }{0.012-x}

given that the value of Ka is small we will ignore -x

x² = 3.162 * 10^{-4} * 0.012

x = 1.948 * 10^{-3}  

Therefore

[ H⁺ ] = 1.948 * 10^{-3}

given that

pH = - Log [ H⁺ ]

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Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

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