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Chapter 14: Problem 70

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

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

Step by step solution

01

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-
02

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+: $$ 6.7 \times 10^{-4} = \frac{[\mathrm{H}^{+}]\frac{x}{3}}{\frac{2x}{3}} $$
03

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: $$ [\mathrm{H}^{+}] = 8.93 \times 10^{-4} \thinspace \mathrm{M} $$
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈦� (base from NaOH) to produce F鈦� and water.
  • This neutralization causes HF to decrease and F鈦� 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鈦篯) 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鈦篯 through equilibrium expressions. First, we use the expression for the acid dissociation constant and rearrange to solve for [H鈦篯:
  • Plug in known concentrations and K_a value into the equation: \[K_a = \frac{[H^+][F^-]}{[HF]}\]
  • Resolve it to find [H鈦篯 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.

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Most popular questions from this chapter

Calculate the pH of a solution prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.20 \mathrm{M}\) ethanolamine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{2}\), with \(50.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{HCl} . \mathrm{K}_{\mathrm{a}}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{3}+\) is \(3.6 \times 10^{-10}\)

A \(0.1375 \mathrm{M}\) solution of potassium hydroxide is used to titrate \(35.00 \mathrm{~mL}\) of \(0.257 M\) hydrobromic acid. (Assume that volumes are additive.) (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of potassium hydroxide is required to reach the equivalence point? (d) What is the \(\mathrm{pH}\) of the solution before any \(\mathrm{KOH}\) is added? (e) What is the \(\mathrm{pH}\) of the solution halfway to the equivalence point? (f) What is the \(\mathrm{pH}\) of the solution at the equivalence point?

Which of the following would form a buffer if added to \(250.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{SnF}_{2} ?\) (a) \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl}\) (b) \(0.060 \mathrm{~mol}\) of \(\mathrm{HCl}\) (c) \(0.040 \mathrm{~mol}\) of \(\mathrm{HCl}\) (d) \(0.040 \mathrm{~mol}\) of \(\mathrm{NaOH}\) (e) \(0.040 \mathrm{~mol}\) of \(\mathrm{HF}\)

WEB A 25.00-mL sample of formic acid, \(\mathrm{HCHO}_{2}\), is titrated with \(39.74 \mathrm{~mL}\) of \(0.117 \mathrm{M} \mathrm{KOH}\). (a) What is \(\left[\mathrm{HCHO}_{2}\right]\) before titration? (b) Calculate \(\left[\mathrm{HCHO}_{2}\right],\left[\mathrm{CHO}_{2}^{-}\right],\left[\mathrm{OH}^{-}\right]\), and \(\left[\mathrm{K}^{+}\right]\) at the equivalence point. (Assume that volumes are additive.) (c) What is the \(\mathrm{pH}\) at the equivalence point?

A solution consisting of \(25.00 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) in \(178 \mathrm{~mL}\) of water is titrated with \(0.114 \mathrm{M}\) KOH. (a) How many \(\mathrm{mL}\) of \(\mathrm{KOH}\) are required to reach the equivalence point? (b) Calculate \(\left[\mathrm{Cl}^{-}\right],\left[\mathrm{K}^{+}\right],\left[\mathrm{NH}_{3}\right]\), and \(\left[\mathrm{OH}^{-}\right]\) at the equivalence point. (Assume that volumes are additive.) (c) What is the \(\mathrm{pH}\) at the equivalence point?
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