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Chapter 15: Problem 36

Calculate the pH of a solution that is 0.60\(M\) HF and 1.00\(M \mathrm{KF}\)

Short Answer

Expert verified
The pH of the solution containing 0.60 M HF and 1.00 M KF is approximately 3.71, using the pKa of HF (\(\approx 3.46\)) and the Henderson-Hasselbalch equation.

Step by step solution

01

Write down the given information

We have a solution containing 0.60 M HF (weak acid) and 1.00 M KF (the source of F-, a conjugate base). Our goal is to calculate the pH of this solution.
02

Find the pKa of HF

The dissociation constant, Ka, of HF is 3.5 脳 10^{-4}. To find the pKa, we use the formula: \(pKa = -\log{Ka}\). So, the pKa of HF is given by: \[pKa = -\log{(3.5 脳 10^{-4})} \approx 3.46\]
03

Use Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pKa + \log{\frac{[A^-]}{[HA]}}\] Where [A-] is the concentration of the conjugate base (F-), and [HA] is the concentration of the weak acid (HF). In our case: \([A^-] = 1.00\,{M}\) (from KF) and \([HA] = 0.60\,{M}\) (HF)
04

Calculate the pH of the solution

Substitute the values of pKa, [A-], and [HA] from Steps 2 and 3 into the Henderson-Hasselbalch equation to find the pH. \[pH = 3.46 + \log{\frac{1.00}{0.60}} \approx 3.46 + 0.25 = 3.71\] So, the pH of the solution containing 0.60 M HF and 1.00 M KF is approximately 3.71.

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is an essential tool for chemists when it comes to calculating the pH of buffer solutions. A buffer solution typically consists of a weak acid and its conjugate base in equilibrium. This equation helps to relate the pH of the solution with the concentrations of the acid and its conjugate base. It is written as follows: \[pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)\] Here,
  • \([A^-]\) stands for the concentration of the conjugate base,
  • \([HA]\) is the concentration of the weak acid,
  • \(pKa\) is the negative logarithm of the acid dissociation constant (\(Ka\)).
The Henderson-Hasselbalch equation elegantly shows how the ratio of the concentrations of the conjugate base and the acid affects the pH, making it crucial for understanding buffer systems.
Weak acid
A weak acid is a type of acid that does not completely dissociate in water. This means that not all of its molecules release hydrogen ions \((H^+)\) into the solution. Common examples include acetic acid and hydrofluoric acid (HF). In the context of pH calculations, weak acids are important because they establish an equilibrium between the un-ionized acid and its conjugate base. Understanding weak acids is essential for calculating the pH of solutions, especially buffers. Their incomplete dissociation results in an equilibrium, which plays a crucial role in buffering changes in pH. Some characteristics of weak acids include:
  • They have higher \(pKa\) values.
  • They can create buffer solutions when mixed with their conjugate bases.
  • The extent of dissociation is important for determining the strength of the acid and its buffering capacity.
Conjugate base
A conjugate base forms when a weak acid loses a hydrogen ion \((H^+)\). It is an integral part of how buffer solutions work. The conjugate base in a buffer solution pairs with the weak acid to resist drastic changes in pH by neutralizing added acids or bases. When a conjugate base is present in solution, it pairs with any excess \(H^+\) ions, limiting the decrease in pH. Likewise, it can stabilize a solution by providing additional \(A^-\) ions if the solution becomes too basic. For example, in our solution:
  • \(F^-\) is the conjugate base that arises when \(HF\) loses a proton \((H^+)\).
  • In the solution of 0.60 M HF and 1.00 M KF, \(KF\) provides the \(F^-\) ions, which is the conjugate base necessary for maintaining pH stability.
This balance between the weak acid and its conjugate base is what makes buffer solutions effective at maintaining steady pH levels.
Dissociation constant (Ka)
The dissociation constant, \(Ka\), is a quantitative measure of the strength of an acid in solution. It reflects the extent to which an acid can donate protons \((H^+)\) to a solution. For a weak acid \((HA)\), the equilibrium can be represented by the equation: \[HA \rightleftharpoons H^+ + A^-\] The dissociation constant is then determined by: \[Ka = \frac{[H^+][A^-]}{[HA]}\] Where:
  • \([H^+]\) is the concentration of hydrogen ions,
  • \([A^-]\) is the concentration of the conjugate base,
  • \([HA]\) is the concentration of the undissociated acid.
The value of \(Ka\) helps us understand how completely an acid dissociates:
  • A large \(Ka\) indicates a stronger acid that can donate protons easily.
  • A smaller \(Ka\) suggests a weaker acid that remains mostly undissociated.
Understanding \(Ka\) is crucial when using the Henderson-Hasselbalch equation, as it directly influences the calculation of \(pKa\), an important factor in determining the pH of buffer solutions.

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Most popular questions from this chapter

Consider the titration of 80.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by 0.400 \(\mathrm{M} \mathrm{HCl}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of HCl have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}} \\ {\text { c. } 30.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}}\end{array} $$

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 \(\mathrm{mL}\) of the weak acid solution has been added to 50.0 \(\mathrm{mL}\) of the 0.100 \(\mathrm{M}\) NaOH solution, the \(\mathrm{pH}\) of the resulting solution is 10.50 . Calculate the original concentration of the solution of weak acid.

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NaOH}\) . The student then adds 13.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) . The pH of the resulting solution is \(4.70 .\) How is the value of \(\mathrm{p} K_{2}\) for the unknown acid related to 4.70\(?\)

The active ingredient in aspirin is acetylsalicylic acid. A 2.51 -g sample of acetylsalicylic acid required 27.36 \(\mathrm{mL}\) of 0.5106 \(\mathrm{M} \mathrm{daOH}\) for complete reaction. Addition of 13.68 \(\mathrm{mL}\) of 0.5106\(M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH \(=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{2}\) value. State any assumptions you must make to reach your answer.

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. $$ \begin{array}{l}{\text { a. } 100.0 \mathrm{mL} \text { of } 0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{HsO}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right) \text { titrated }} \\ {\text { by } 0.10 \mathrm{M} \mathrm{NaOH}}\end{array} $$ $$ \begin{array}{l}{\text { b. } 100.0 \mathrm{mL} \text { of } 0.10 M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right) \text { titrated }} \\ {\text { by } 0.20 \mathrm{M} \mathrm{HNO}_{3}}\end{array} $$ $$ 100.0 \mathrm{mL} \text { of } 0.50 \mathrm{M} \mathrm{HCl} \text { titrated by } 0.25 \mathrm{M} \mathrm{NaOH} $$
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