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

C = pK
C = 4.4E-4*0.032 = about 1.41E-5

H2CO3 ==> H^+ + HCO3^-

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

Answer Link

aliasger2709 aliasger2709 · Answer 2 · 2022-03-09T00:58:11-05:00

pH is the potential of the hydrogen ion concentration in the solution and depicts the acidic and basic behaviour of the solution. 6.27 is the pH of the water at equilibrium.

What is acid dissociation constants?

Acid dissociation constants are the quantity that depicts the acidic strength of the solution and is given by the concentration of the conjugate base, hydrogen ions and the chemical species.

The reaction between carbon dioxide and water can be shown as,

\(\rm CO_{2} + H_{2}O \leftrightharpoons H^{+} + HCO_{3}^{-}\)

Given,

\(\rm K_{CO_{2}}\) = \(0.032 \rm M/\;atm\)

\(\rm P_{CO_{2}}\) = \(4.4 \times 10^{-4}\;\rm atm\)

The amount of the carbon dioxide dissolved can be estimated as,

\(\begin{aligned}\rm CO_{2} &= \rm K_{CO_{2}} \times P_{CO_{2}} \\\\&= 0.032 \rm M/\;atm \times 4.4 \times 10^{-4}\;\rm atm\\\\&= 0.01408 \times 10^{-5}\end{aligned}\)

Acid dissociation can be shown as,

\(\rm K_{a} = \dfrac{[H^{+}]^{2}}{[CO_{2}]}\)

Here, \(\rm K_{a} = 4.46 \times 10^{-7}\)

Substituting value in the equation:

\(\begin{aligned} 4.46 \times 10^{-7} &= \rm \dfrac{[H^{+}]^{2}}{0.01408 \times 10^{-5}}\\\\\rm [H^{+}]^{2}&= 0.2800 \times 10^{-12}\\\\&= 0.529 \times 10^{-6}\end{aligned}\)

From this, pH can be calculated as,

\(\begin{aligned}\rm pH &= \rm -log [H^{+}]\\\\&= \rm -log ( 0.529 \times 10^{-6})\\\\&= 6.27\end{aligned}\)

Therefore, 6.27 is the pH of the water at equilibrium.

Learn more about pH and acid dissociation here:

https://brainly.com/question/8120055