Respuesta :
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.
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.
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,
[tex]\rm CO_{2} + H_{2}O \leftrightharpoons H^{+} + HCO_{3}^{-}[/tex]
Given,
[tex]\rm K_{CO_{2}}[/tex] = [tex]0.032 \rm M/\;atm[/tex]
[tex]\rm P_{CO_{2}}[/tex] = [tex]4.4 \times 10^{-4}\;\rm atm[/tex]
The amount of the carbon dioxide dissolved can be estimated as,
[tex]\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}[/tex]
Acid dissociation can be shown as,
[tex]\rm K_{a} = \dfrac{[H^{+}]^{2}}{[CO_{2}]}[/tex]
Here, [tex]\rm K_{a} = 4.46 \times 10^{-7}[/tex]
Substituting value in the equation:
[tex]\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}[/tex]
From this, pH can be calculated as,
[tex]\begin{aligned}\rm pH &= \rm -log [H^{+}]\\\\&= \rm -log ( 0.529 \times 10^{-6})\\\\&= 6.27\end{aligned}[/tex]
Therefore, 6.27 is the pH of the water at equilibrium.
Learn more about pH and acid dissociation here:
https://brainly.com/question/8120055
