Question

Solve an equilibrium problem (using an ICE table) to calculate the \(\mathrm{pH}\) of each solution. a. \(0.15 \mathrm{MHF}\) b. 0.15 M NaF c. a mixture that is \(0.15 \mathrm{M}\) in HF and \(0.15 \mathrm{M}\) in NaF

Short answer

To find the pH for the solutions: a) For 0.15 M HF, set up an ICE table and solve for H+ concentration then calculate pH. b) For 0.15 M NaF, set up an ICE table for hydrolysis, find OH- concentration, then convert to pH. c) For the mixture, consider the common ion effect, set up an ICE table with the new initial concentrations of HF and F- from NaF, solve for the new H+ concentration, and calculate pH.

Step-by-step solution

Write Down the Equilibrium Expression for HF

HF dissociates in water to form H+ and F- ions. The equilibrium expression is: HF ⇌ H+ + F-The equilibrium constant (Ka) for HF is needed to solve for the concentration of H+ ions, and thus the pH. The Ka value for HF is around 3.5 x 10^-4.

Create an ICE Table for 0.15 M HF

Setup an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations:Initial: [HF] = 0.15 M, [H+] = 0, [F-] = 0Change: [HF] decreases by x, [H+] increases by x, [F-] increases by xEquilibrium: [HF] = 0.15 - x, [H+] = x, [F-] = xUse the Ka expression to solve for x:Ka = [H+][F-]/[HF]Ka = x*x / (0.15 - x)3.5 x 10^-4 = x^2 / (0.15 - x)

Solve for x (concentration of H+ ions)

Assume that x is much less than 0.15 since HF is a weak acid, simplifying the expression to:3.5 x 10^-4 = x^2 / 0.15Solve for x to get the concentration of H+.

Calculate pH for 0.15 M HF

pH is the negative logarithm (base 10) of the H+ concentration. pH = -log[H+]. Use the value of x from Step 3 to calculate the pH.

Write Down the Dissociation Expression for NaF

NaF is a salt that completely dissociates in water to form Na+ and F- ions. The F- ions will be at a concentration of 0.15 M. Since NaF is a salt derived from a weak acid (HF) and a strong base (NaOH), it will undergo hydrolysis in water. The expression for hydrolysis is: F- + H2O ⇌ HF + OH-.

Create an ICE Table for 0.15 M NaF

Setup an ICE table for F- hydrolysis:Initial: [F-] = 0.15 M, [OH-] = 0Change: [F-] decreases by x, [OH-] increases by xEquilibrium: [F-] = 0.15 - x, [OH-] = xUsing Kb (the base dissociation constant) which is related to Ka by the equation Kb = Kw / Ka, where Kw is the ion product of water (1.0 x 10^-14 at 25°C), calculate the hydrolysis of F-.

Solve for x (concentration of OH- ions) for NaF

Solve the hydrolysis expression to find x, which is the concentration of OH- ions.Kb = [OH-][HF]/[F-]Kb = x*x / (0.15 - x)Calculate Kb using Ka and Kw, and then solve for x.

Calculate pOH and pH for 0.15 M NaF

pOH is the negative logarithm (base 10) of the OH- concentration (pOH = -log[OH-]) and pH + pOH = 14. Use the pOH and this relationship to find the pH.

Analyze the Mixture of HF and NaF

In a mixture of 0.15 M HF and 0.15 M NaF, the F- ions from NaF will shift the equilibrium of the dissociation of HF to the left, according to Le Chatelier's principle. This will result in a decrease in [H+] released from HF. The solution will already have [F-] = 0.15 M from the NaF so the ICE table will need to be modified. This will create a common ion effect scenario. The calculation for pH will take the additional source of F- into account.

Solve for the New [H+] in the Mixture

With the modified ICE table, solve for the new concentration of H+ ions in the presence of the common ion from NaF. Since [F-] is no longer zero due to NaF, the expression will be: Ka = [H+][F-]/([HF] - [H+])Since [HF] and [F-] start at 0.15 M, the expression simplifies to find the new [H+].

Calculate the pH of the Mixture

With the new [H+] from Step 10, calculate the pH in the same way as before, using the pH = -log[H+].

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is crucial when tackling problems involving the calculation of pH using an ICE table. An equilibrium constant, represented as Ka for acids or Kb for bases, quantifies the extent of dissociation of a compound in water. For a weak acid like HF, the expression is written as Ka = [H⁺][F^-]/[HF], where the concentrations are those at equilibrium. By setting up an ICE table, we can delineate the initial concentrations, the changes that occur as the system reaches equilibrium, and the final equilibrium concentrations. Solving for x in the equation gives the concentration of H⁺ ions and subsequently allows us to find the pH.

Weak Acid Dissociation

Weak acids, such as HF, only partially dissociate in water, establishing an equilibrium between the undissociated acid and the ions formed. When we express the dissociation of HF, we write HF ⇌ H⁺ + F^-. To calculate the pH, we start with an ICE table. The initial concentration of the acid is given, and initially, there are no H⁺ or F^- ions. Upon dissociation, the concentration of HF decreases by x, while H⁺ and F^- increase by x. The equilibrium constant expression comes into play as we solve for x, which corresponds to the concentration of H⁺ ions produced, allowing us to then calculate the pH of the solution.

Hydrolysis of Salts

The hydrolysis of salts is another factor that can affect the pH of a solution. For example, when NaF dissolves in water, it completely dissociates into Na⁺ and F^- ions. However, the F^- ion, being the conjugate base of the weak acid HF, will react with water to produce HF and OH^- ions according to the reaction F^- + H₂O ⇌ HF + OH^-. Here, we can set up an ICE table for hydrolysis to find the concentration of OH^-, which can be used to determine the pOH and then the pH. It's important to remember that the hydrolysis equation for the salt should be aligned with the equilibrium expression for Kb, and the relationship between Ka, Kb, and the ion product of water, Kw, to ensure accurate pH calculations.

Common Ion Effect

Finally, let's consider the common ion effect, which arises when a solution contains two substances that share a common ion. In the mixture of HF and NaF, both contain the common ion F^-. According to Le Chatelier's principle, the presence of additional F^- from NaF will suppress the dissociation of HF, effectively reducing the concentration of H⁺ ions and shifting the equilibrium to the left. This effect must be taken into account when creating an ICE table for the mixture. The modified equilibrium constant expression helps us determine the concentration of H⁺ in the presence of the common ion, leading to an accurate calculation of the mixture's pH. This demonstrates how the common ion effect can stabilize the pH of a solution by preventing excessive dissociation of a weak acid or base.