Answer:
pH=2.82
Explanation:
pH is a scale used to indicate if a water-based solution is acidic or basic, and is defined by the following equation:
[tex]pH=- log (H^{+})[/tex] Equation (1)
In aqueous solutions there are no ions [tex]H^{+}[/tex], because the proton is transferred to [tex]H_{2}O[/tex] to form hydronium ions, [tex]H_{3}O^{+}[/tex]. Thus, we can determine the pH of the solution, finding the molar concentration of [tex]H_{3}O^{+}[/tex] ions:
[tex][pH=- log (H_{3}O^{+})[/tex] Equation (2)
The solution prepared contains two weak acids, each one undergoes an equilibrium process with the corresponding equilibrium constant Ka, as it is represented in the following reaction for a weak acid HA:
HA + [tex]H_{2}O[/tex] ⇔ [tex]H_{3}O^{+}[/tex] + [tex]A^{-}[/tex]
The equilibrium constant is defined by the equilibrium concentration of products over reactants:
[tex]k= \frac{[H_3O^{+}][A^{-}]}{[HA][H_2O]}[/tex] Equation (3)
However, the molar concentration of water is essentially constant for reactions in aqueous solutions, then the acid dissociation constant is defined as follow:
[tex]k_a=K[H_2O]= \frac{[H_3O^{+}][A^{-}]}{[HA]}[/tex] Equation (4)
Information for [tex]CH_3COOH[/tex]:
Reaction: [tex]CH_3COOH[/tex] + [tex]H_{2}O[/tex]⇔ [tex]H_{3}O^{+}[/tex] + [tex]CH_3COO^{-}[/tex]
Initial moles per Liter: 0.125 M + A ⇔ 0 + 0
Reacting moles per liter: -x + -x ⇔ x + x
Equilibrium mole per liter: 0.125 M-x + A-x ⇔ x + x
[tex]k_a= 1.8*10^{-5}[/tex]
Information for [tex]H_3BO_3[/tex]:
Reaction: [tex]H_3BO_3[/tex] + [tex]H_{2}O[/tex] ⇔ [tex]H_{3}O^{+}[/tex] + [tex]H_2BO_3^{-}[/tex]
Initial moles per Liter: 0.125 M + A ⇔ 0 + 0
Reacting moles per liter: -y + -y ⇔ y + y
Equilibrium mole per liter: 0.125 M-y + A-y ⇔ y + y
[tex]k_a= 5.4*10^{-10}[/tex]
Replacing the equilibrium information of each acid in equation 4, we get:
For [tex]CH_3COOH[/tex]:
[tex]1.8*10^{-5} =\frac{[x][x]}{[0.125-x]}[/tex] Equation (5)
For [tex]H_3BO_3[/tex]:
[tex]5.4*10^{-10} =\frac{[y][y]}{[0.125-y]}[/tex] Equation (6)
Solving equation 5 and 6:
For [tex]CH_3COOH[/tex]:
[tex]x = [H_{3}O^{+}] = 1.49*10^{-3}[/tex]
For [tex]H_3BO_3[/tex]:
[tex]y = [H_{3}O^{+}] = 1.64*10^{-6}[/tex]
Due to both acids are in the same solution, the total concentration of [tex][H_{3}O^{+}][/tex] is (x+y) [tex]1.5064*10^{-3}[/tex]. Replacing this concentration in equation 2, we get:
[tex]pH= -log (1.5064*10^{-3})=2.82[/tex]
