Question

Weak acids and bases are those that do not completely dissociate in water. The dissociation of the acid or base is an equilibrium process and has a corresponding equilibrium constant. Ka is the equilibrium constant for the dissociation of a weak acid and Kb is the equilibrium constant for the dissociation of a weak base. What is the pH of a solution that has 0.125 M CH3COOH and 0.125 M H3BO3? Ka of CH3COOH = 1.8 × 10−5 and Ka of H3BO3 = 5.4 × 10−10 ANSWER Unselected 4.74 Unselected 5.64 Unselected 2.82 Unselected 2.52 Unselected

Answer 1

pH=2.82

Step-by-step explanation:

pH is a scale used to indicate if a water-based solution is acidic or basic, and is defined by the following equation:

`pH=- log (H^(+))` Equation (1)

In aqueous solutions there are no ions
`H^(+)`, because the proton is transferred to
`H_(2)O` to form hydronium ions,
`H_(3)O^(+)`. Thus, we can determine the pH of the solution, finding the molar concentration of
`H_(3)O^(+)` ions:

`[pH=- log (H_(3)O^(+))` 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 +
`H_(2)O` ⇔
`H_(3)O^(+)` +
`A^(-)`

The equilibrium constant is defined by the equilibrium concentration of products over reactants:

`k= ([H_3O^(+)][A^(-)])/([HA][H_2O])` 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:

`k_a=K[H_2O]= ([H_3O^(+)][A^(-)])/([HA])` Equation (4)

Information for
`CH_3COOH`:

Reaction:
`CH_3COOH` +
`H_(2)O`⇔
`H_(3)O^(+)` +
`CH_3COO^(-)`

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

`<em><strong>k_a=</strong></em> 1.8*10^(-5)`

Information for
`H_3BO_3`:

Reaction:
`H_3BO_3` +
`H_(2)O` ⇔
`H_(3)O^(+)` +
`H_2BO_3^(-)`

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

`<em><strong>k_a=</strong></em> 5.4*10^(-10)`

Replacing the equilibrium information of each acid in equation 4, we get:

For
`CH_3COOH`:

`1.8*10^(-5) =([x][x])/([0.125-x])` Equation (5)

For
`H_3BO_3`:

`5.4*10^(-10) =([y][y])/([0.125-y])` Equation (6)

Solving equation 5 and 6:

For
`CH_3COOH`:

`x = [H_(3)O^(+)] = 1.49*10^(-3)`

For
`H_3BO_3`:

`y = [H_(3)O^(+)] = 1.64*10^(-6)`

Due to both acids are in the same solution, the total concentration of
`[H_(3)O^(+)]` is (x+y)
`1.5064*10^(-3)`. Replacing this concentration in equation 2, we get:

`pH= -log (1.5064*10^(-3))=<em><strong>2.82</strong></em>`