Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
The pH of the solution that is 0.60 M HF and 1.00 M KF is approximately 3.40.

Step by step solution

01

Find the pKa value for HF

Use the formula: \( pKa = -\log{Ka} \) Using the given Ka value for HF: \( pKa = -\log{(6.6 × 10^{-4})} \) Calculating the pKa value: \( pKa \approx 3.18 \)
02

Apply the Henderson-Hasselbalch equation

Use the formula: \( pH = pKa + \log{\frac{[A^-]}{[HA]}} \) Substitute the values we found in step 1 and the given concentrations: \( pH = 3.18 + \log{\frac{1.00}{0.60}} \) Simplify and calculate the pH value: \( pH = 3.18 + \log{1.67} \) \( pH \approx 3.18 + 0.22 \) \( pH \approx 3.40 \) So, the pH of the solution that is 0.60 M HF and 1.00 M KF is approximately 3.40.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Acid-Base Chemistry
Acid-base chemistry is a fundamental area of chemistry that explores how acids and bases interact in a solution. ### Understanding Acids and Bases In simple terms, acids are substances that can donate a proton ( H^+ ) to another substance, while bases are substances that can accept a proton. This interaction is crucial for many chemical reactions. * **Acid Example**: Hydrofluoric acid (HF) is an example of a weak acid, meaning it does not completely dissociate in water. * **Base Example**: Potassium fluoride (KF), on the other hand, acts as a base when it dissociates to form fluoride ions (F^-) in solution. ### Importance of Acid and Base Strength The strength of an acid or base affects how it behaves in a reaction. In this case, HF is a weak acid, which means only a small fraction of HF molecules donate their protons. The numerical constant that helps to describe this tendency is the acid dissociation constant (Ka), and its negative logarithm, pKa, is crucial for calculating pH with the Henderson-Hasselbalch equation.
pH Calculation
Calculating the ph of a solution involves understanding the concentration of hydrogen ions (H^+) in a solution. * **Formula**: The most common formula used for pH calculation is \( pH = -\log{[H^+]} \).In the specific exercise provided, we use a different approach called the Henderson-Hasselbalch equation, which is especially useful for buffer solutions. ### Henderson-Hasselbalch EquationThe Henderson-Hasselbalch equation is \( pH = pKa + \log{\frac{[A^-]}{[HA]}} \).* **pKa**: This is the negative log of the acid dissociation constant. For HF, \( pKa \approx 3.18 \).* **[A^-] and [HA]**: These represent the concentrations of the conjugate base (F^-) and the acid (HF), respectively.Inserting these values helps you calculate the pH effectively, especially when dealing with a buffer solution containing both the acid and its conjugate base.
Buffer Solution
A buffer solution is a special type of solution that resists changes in ph when small amounts of acid or base are added. ### How Buffers Work Buffers typically contain a weak acid and its conjugate base, or vice versa. They maintain pH stability by neutralizing added acids or bases. ### Components of a Buffer System For the given problem, the buffer solution is made from: * **Weak Acid**: HF (Hydrofluoric Acid) serves as the weak acid. * **Conjugate Base**: KF, which provides the F^- ions as the conjugate base. ### Why Buffers are Important Buffers are crucial in many biological and chemical processes. They help maintain the pH level necessary for reactions, ensuring that processes like enzyme function and nutrient absorption can proceed efficiently. Using the Henderson-Hasselbalch equation, you get an understanding of how the concentration of acid and base in the buffer solution impacts the overall pH. This stability is vital in many practical applications, from biological systems to industrial processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=\) \(7.4 \times 10^{-5}\). Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=1.0 M ?\)

Consider the following four titrations. i. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HCl}\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) ii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaOH}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iv. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HF titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing \(\mathrm{pH}\) initially before any titrant has been added. c. increasing \(\mathrm{pH}\) at the halfway point in equivalence. d. increasing \(\mathrm{pH}\) at the equivalence point. How would the rankings change if \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) replaced \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and if \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) replaced \(\mathrm{HF}\) ?

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}}\), would you add \(\mathrm{HCl}\) or \(\mathrm{NaOH}\) ? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{~L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) ?

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\). b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the \(\mathrm{pH}\). c. Do the same as in part b, but use the following equilibrium to calculate the \(\mathrm{pH}\) : \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q)\) d. Do your answers in parts \(\mathrm{b}\) and \(\mathrm{c}\) agree? Explain.

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free