Question

Using data from Appendix D, calculate [OH \(^{-}\) ] and pH for each of the following solutions: (a) \(0.105 \mathrm{M}\) NaF, (b) \(0.035 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S},(\mathrm{c})\) a mixture that is \(0.045 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(0.055 \mathrm{M}\) in \(\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{Ba} .\)

Short answer

For the given solutions, the [OH⁻] and pH values are: (a) 0.105 M NaF: [OH⁻] = \( 1.23 \times 10^{-6} \) M; pH ≈ 8.09 (b) 0.035 M Na₂S: [OH⁻] = \( 2.19 \times 10^{-4} \) M; pH ≈ 10.34 (c) 0.045 M CH₃COONa and 0.055 M (CH₃COO)₂ Ba: [OH⁻] = \( 1.95 \times 10^{-5} \) M; pH ≈ 9.29

Step-by-step solution

(a) Calculate [OH⁻] and pH for 0.105 M NaF solution

1. Finding the Kb of F⁻: The anion of NaF is F⁻, which acts as a weak base. Its conjugate acid is HF with a known Ka. First, we need to find the Kb of F⁻ using the relation between Ka and Kb: Ka * Kb = Kw. Here, Kw is the ion product of water, 1x 10^{-14}. Ka of HF = 6.8 x 10^{-4} Kb = Kw / Ka = \( \frac{1 \times 10^{-14}}{6.8 \times 10^{-4}} \) = \( 1.47 \times 10^{-11} \) 2. Set up an ICE table for the hydrolysis reaction of F⁻: F⁻ + H₂O ⇌ HF + OH⁻ Initial: 0.105 M - - 0 Change: -x +x +x Equilibrium: 0.105-x x x 3. Solve for x ([OH⁻]): Kb = [HF][OH⁻]/[F⁻] = \( \frac{x \times x}{0.105 - x} \) = 1.47 x 10^{-11} Since the concentration of F⁻ is much greater than Kb, we can assume x is much smaller than 0.105 and ignore it in the denominator: x squared ≈ 0.105 x (1.47 x 10^{-11}) x ≈ [OH⁻]= 1.23 x 10^{-6} M 4. Calculate the pOH and pH: pOH = -log[OH⁻] = -log(1.23 x 10^{-6}) ≈ 5.91 pH = 14 - pOH = 14 - 5.91 ≈ 8.09

(b) Calculate [OH⁻] and pH for 0.035 M Na₂S solution

1. Finding the Kb of S²⁻: S²⁻ is the anion in Na₂S, which acts as a weak base. Its conjugate acid is HS⁻, with a known Ka. First, we need to find the Kb of S²⁻ using the relation between Ka and Kb: Ka * Kb = Kw. Here, Kw is the ion product of water, 1x 10^{-14}. Ka of HS⁻ = 1.3 x 10^{-7} Kb = Kw / Ka = \( \frac{1 \times 10^{-14}}{1.3 \times 10^{-7}} \) = \( 7.69 \times 10^{-8} \) 2. Set up an ICE table for the hydrolysis reaction of S²⁻: S²⁻ + H₂O ⇌ HS⁻ + OH⁻ Initial: 0.035 M - - 0 Change: -x +x +x Equilibrium: 0.035-x x x 3. Solve for x ([OH⁻]): Kb = [HS⁻][OH⁻]/[S²⁻] = \( \frac{x \times x}{0.035 - x} \) = 7.69 x 10^{-8} Since the concentration of S²⁻ is much greater than Kb, we can assume x is much smaller than 0.035 and ignore it in the denominator: x squared ≈ 0.035 x (7.69 x 10^{-8}) x ≈ [OH⁻] = 2.19 x 10^{-4} M 4. Calculate the pOH and pH: pOH = -log[OH⁻] = -log(2.19 x 10^{-4}) ≈ 3.66 pH = 14 - pOH = 14 - 3.66 ≈ 10.34

(c) Calculate [OH⁻] and pH for a mixture that is 0.045 M CH₃COONa and 0.055 M (CH₃COO)₂Ba solution

1. Calculate the total concentration of CH₃COO⁻: Since both salts share the same anion, CH₃COO⁻, we can find the total concentration of CH₃COO⁻: [CH₃COO⁻] = 0.045 + 2 * 0.055 = 0.155 M 2. Finding the Ka of CH₃COOH and Kb of CH₃COO⁻: Ka of CH₃COOH = 1.8 x 10^{-5} Kb of CH₃COO⁻ = Kw / Ka = \( \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} \) = \( 5.56 \times 10^{-10} \) 3. Set up an ICE table for the hydrolysis reaction of CH₃COO⁻: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ Initial: 0.155 M - - 0 Change: -x +x +x Equilibrium: 0.155-x x x 4. Solve for x ([OH⁻]): Kb = [CH₃COOH][OH⁻]/[CH₃COO⁻] = \( \frac{x \times x}{0.155 - x} \) = 5.56 x 10^{-10} Since the concentration of CH₃COO⁻ is much greater than Kb, we can assume that x is much smaller than 0.155 and ignore it in the denominator: x squared ≈ 0.155 x (5.56 x 10^{-10}) x ≈ [OH⁻] = 1.95 x 10^{-5} M 5. Calculate the pOH and pH: pOH = -log[OH⁻] = -log(1.95 x 10^{-5}) ≈ 4.71 pH = 14 - pOH = 14 - 4.71 ≈ 9.29

Key concepts

Hydrolysis Reaction

In the realm of chemistry, a hydrolysis reaction is a pivotal process which typically occurs when a salt, made from a weak acid or base, dissolves in water. In simpler terms, hydrolysis involves a salt reacting with water to form an acid or a base. For instance, when sodium fluoride (NaF) is dissolved in water, the fluoride ion (F⁻) undergoes hydrolysis with water to yield hydroxide ions (OH⁻) and hydrofluoric acid (HF).

This reaction can be represented as:
F⁻ + H₂O ⇌ HF + OH⁻.

Understanding the behavior of these ions in water is crucial for calculating pH. The generated OH⁻ increases the solution's basicity, thus affecting the overall pH level.

Ionic Product of Water (Kw)

The ionic product of water, or Kw, is an essential constant in the field of acid-base chemistry. It signifies the product of the molar concentrations of H⁺ and OH⁻ ions in pure water at a given temperature. Generally, at 25°C, Kw equals 1 × 10⁻¹⁴.

The equation can be expressed as:
Kw = [H⁺][OH⁻].

This constant is foundational because it connects the concentrations of hydrogen and hydroxide ions, which are pivotal for understanding the pH and pOH of solutions. Additionally, it provides a relation between the acid dissociation constant (Ka) and base dissociation constant (Kb) for conjugate acid-base pairs: Ka × Kb = Kw.

ICE Table Method

The ICE Table Method is an organized way to solve for concentrations of reactants and products in an equilibrium reaction. ICE stands for Initial, Change, and Equilibrium. You can depict the concentrations at each of these stages for all entities in the reaction.

For instance, consider the fluoride ion hydrolysis described earlier. The ICE table would be:

Initial: [F⁻] = 0.105 M, [HF] = 0, [OH⁻] = 0
Change: [F⁻] = −x, [HF] = +x, [OH⁻] = +x
Equilibrium: [F⁻] = (0.105 - x) M, [HF] = x, [OH⁻] = x

In this method, 'x' represents the change in concentration. This method significantly simplifies the process of finding equilibrium concentrations in acid-base reactions and thus assists in pH calculations.

Acid-Base Equilibrium

Acid-base equilibrium pertains to the balance that exists in a solution between its acid (H⁺) and base (OH⁻) components. It is crucial for the determination of pH, as it touches on the extents to which acids donate hydrogen ions or bases accept them. The equilibrium lies at the core of the calculations involving the ICE table method.

When solving for pH, one calculates the equilibrium concentration of OH⁻ by setting up an equation with the base dissociation constant (Kb), which can be derived from the ionic product of water (Kw) and the acid dissociation constant (Ka). In summary, acid-base equilibrium is indispensable for accurately determining the acidity or basicity of a solution.