Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 25

You are asked to prepare a pH \(=2.50\) buffer solution starting from $1.50 \mathrm{~L}\( of a \)0.75 \mathrm{M}$ solution of hydrofluoric acid (HF) and any amount you need of sodium fluoride (NaF). (a) What is the \(\mathrm{pH}\) of the hydrofluoric acid solution prior to adding sodium fluoride? (b) How many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added.

Short Answer

Expert verified
The initial pH of the 0.75 M hydrofluoric acid solution is approximately 1.72. To prepare the desired pH 2.50 buffer solution, we need to add approximately 1.97 grams of sodium fluoride to the 1.50 L of 0.75 M hydrofluoric acid solution.

Step by step solution

01

1. Ionization equation and ionization constant for hydrofluoric acid

Hydrofluoric acid (HF) is a weak acid and ionizes in water to form hydrogen ions (H+) and fluoride ions (F-): \[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\] For weak acids, the equilibrium ionization constant (Ka) is given by: \[K_a = \frac{[H^+][F^-]}{[HF]}\]
02

2. Calculate the starting pH of the solution

To calculate the pH of the 0.75 M hydrofluoric acid solution, we use the Ka expression given below: \[ K_a = \frac{x^2}{0.75-x}\] The ionization constant of HF is approximately \(K_a = 6.6 \times 10^{-4}\). Since HF is a weak acid, we can assume \[ x \ll 0.75\] Therefore, \[ 6.6 \times 10^{-4} \approx \frac{x^2}{0.75}\] Solving for x: \[ x = \sqrt{6.6 \times 10^{-4} \times 0.75} = 0.01894\] Here, x represents the concentration of hydrogen ions [H+]. Thus, the initial pH of the HF solution can be determined using the formula: \[pH = -\log[H^+]\] \[ pH = -\log(0.01894) \approx 1.72\] The initial pH of the hydrofluoric acid solution is approximately 1.72.
03

3. Determine the required concentration of fluoride ions

We are asked to prepare a pH 2.50 buffer solution using HF and NaF. We will use the expression for Ka and the desired pH to determine the concentration of fluoride ions (F-) needed in the solution: \[ K_a = \frac{[H^+][F^-]}{[HF]}\] We know \([H^+] = 10^{-pH} = 10^{-2.5}\), and the final [HF] should remain constant at 0.75 M, as the volume change is negligible: \[ 6.6 \times 10^{-4} = \frac{(10^{-2.5})([F^-])}{0.75}\] Solve for [F-]: \[ [F^-] = \frac{6.6 \times 10^{-4} \times 0.75}{10^{-2.5}} \approx 0.0312\mathrm{M}\]
04

4. Calculate the amount of sodium fluoride needed

Now we know the required concentration of fluoride ions, we can calculate the amount of sodium fluoride, NaF, needed to be added: Moles of F- ions needed: \[moles = M_{required} \times V_{solution} = 0.0312 \times 1.50 = 0.0468 \: mol\] To find the mass of NaF required, we multiply the moles by the molar mass (42.0 g/mol for NaF): \[mass = moles \times molar \: mass = 0.0468 \times 42.0 = 1.97\mathrm{~g}\] To prepare the desired pH 2.50 buffer solution, we need to add approximately 1.97 grams of sodium fluoride to the 1.50 L of 0.75 M hydrofluoric acid solution.

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.

Start your free trial

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

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

How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

Predict whether the equivalence point of each of the following titrations is below, above, or at pH 7: (a) benzoic acid titrated with KOH, (b) ammonia titrated with iodic acid, (c) hydroxylamine with hydrochloric acid.

Consider the equilibrium $$ \mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the pH of the solution increase, decrease, or stay the same?

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating $0.200 \mathrm{M}\( solutions of each of the following bases with 0.200 \)M$ HBr: \((\mathbf{a})\) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

A 10.0-mL sample of \(0.250 \mathrm{M}\) acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\( is titrated with \)0.100 \mathrm{M}$ KOH solution. Calculate the pH after the following volumes of base have been added: (a) \(0 \mathrm{~mL},\) (b) \(12.5 \mathrm{~mL}\) (c) \(24.5 \mathrm{~mL}\) (d) \(25.0 \mathrm{~mL}\) (e) \(25.5 \mathrm{~mL}\) (f) \(30.0 \mathrm{~mL}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation
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