Question

Consider the titration of \(\mathrm{HF}\left(K_{\mathrm{a}}=6.7 \times 10^{-4}\right)\) with \(\mathrm{NaOH}\). What is the \(\mathrm{pH}\) when a third of the acid has been neutralized?

Short answer

Answer: The pH of the solution is approximately 3.05 when one-third of the acid has been neutralized.

Step-by-step solution

Determine the Moles of HF and F- after Neutralization

Let's assume that the initial number of moles of HF is x. After one-third of the moles of HF have been neutralized by NaOH, we will have 2x/3 moles of HF and x/3 moles of F-

Setting up the Equation for H+ Concentration

Now, we can set up the equation for the H+ concentration. The acid dissociation constant, Ka, is given by the following expression: $$ K_a = \frac{[\mathrm{H}^{+}][\mathrm{F}^{-}]}{[\mathrm{HF}]} $$ We are given the value of Ka for HF, which is \(6.7 \times 10^{-4}\). From step 1, we know the concentrations of HF and F-. Let's substitute these values into the Ka expression to find the concentration of H+: $$ 6.7 \times 10^{-4} = \frac{[\mathrm{H}^{+}]\frac{x}{3}}{\frac{2x}{3}} $$

Solve for H+ Concentration

Now, we can solve the equation for the H+ concentration. $$ [\mathrm{H}^{+}] = 6.7 \times 10^{-4} \times \frac{\frac{2x}{3}}{\frac{x}{3}} = \frac{4}{3}(6.7 \times 10^{-4}) $$ After calculating the value, we get: $$ [\mathrm{H}^{+}] = 8.93 \times 10^{-4} \thinspace \mathrm{M} $$

Calculate the pH of the Solution

Finally, we can use the H+ concentration to calculate the pH of the solution using the following formula: $$ \mathrm{pH} = -\log[\mathrm{H}^{+}] $$ Substituting the H+ concentration, we get: $$ \mathrm{pH} = -\log(8.93 \times 10^{-4}) $$ After calculating, the pH of the solution is about 3.05 when one-third of the acid has been neutralized.

Key concepts

Acid-Base Equilibrium

Acid-base equilibrium involves the balance between acid and base species in a solution. In our titration problem, we're dealing with hydrofluoric acid (HF) and sodium hydroxide (NaOH). As NaOH, a strong base, is added to the HF solution, it starts converting HF molecules into their conjugate base, the fluoride ion ( F^- ). This is how the equilibrium shifts:

• HF (acid) reacts with OH⁻ (base from NaOH) to produce F⁻ and water.
• This neutralization causes HF to decrease and F⁻ to increase.

Such changes are governed by the Law of Mass Action, which states that the reaction will adjust to maintain an equilibrium constant, known as the acid dissociation constant, ( K_a ). Understanding these shifts is crucial for solving the titration equation, especially when calculating concentrations needed in further steps.

pH Calculation

The pH value represents how acidic or basic a solution is. It's calculated from the concentration of hydrogen ions ([H⁺]) in the solution using the formula: \[\text{pH} = -\log[H^+]\]In the given exercise, after a portion of HF is neutralized by NaOH, we solve for [H⁺] through equilibrium expressions. First, we use the expression for the acid dissociation constant and rearrange to solve for [H⁺]:

• Plug in known concentrations and K_a value into the equation: \[K_a = \frac{[H^+][F^-]}{[HF]}\]
• Resolve it to find [H⁺] using simple algebraic manipulations.
• This ultimately gives [H^+] = 8.93 \times 10^{-4} \thinspace M.

By calculating \(-\log(8.93 \times 10^{-4})\), you determine the solution's pH to be about 3.05. A lower pH indicates the presence of more H^+, typical of acidic solutions.

Acid Dissociation Constant

The acid dissociation constant, K_a, is essential for understanding an acid's strength in water. It quantifies how readily an acid donates a proton to form its conjugate base. In our HF titration example, K_a is critical for calculating hydrogen ion concentration:

• K_a indicates how much HF dissociates into H^+ and F^- in solution: \[K_a = \frac{[H^+][F^-]}{[HF]}\]
• A higher K_a means a stronger acid because more of the acid is dissociated.
• Here, HF has a K_a of 6.7 \times 10^{-4}, indicating it's weaker than strong acids like HCl but not negligible.

Using K_a helps predict pH changes during neutralization in titration. Recognizing K_a's role lets you solve equilibrium problems, tying together concepts of acid strength and pH calculation.