Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 33

By what factor does \(\left[\mathrm{H}^{+}\right]\) change for a pH change of (a) 2.00 units, (b) 0.50 units?

Short Answer

Expert verified
For a pH change of (a) 2.00 units, the factor by which \(\mathrm{[H^+]}\) changes is \(10^2\) or 100 times. For a pH change of (b) 0.50 units, the factor by which \(\mathrm{[H^+]}\) changes is \(10^{0.5}\) or approximately 3.16 times.

Step by step solution

01

Recall the definition of pH

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen ion concentration: \(pH = -\log_{10}{\mathrm{[H^+]}}\)
02

Write the pH change equation

Given a change in pH, we want to find the corresponding change in the hydrogen ion concentration. So if the pH changes by \(x\) units, we can write the equation as: \(\Delta pH = pH_2 - pH_1 = x\)
03

Calculate the Factor of Change

To find the factor by which \(\mathrm{[H^+]}\) changes, we first need to express the hydrogen ion concentrations for the two different pH values as powers of 10. By doing so, we'll be able to establish the factor of change between them.
04

Step 3(a): Calculate the factor of change for 2.00 units of pH change

Given \(\Delta pH = 2.00\), let's find the hydrogen ion concentration at the two pH values: \(pH_1 = -\log_{10}{\mathrm{[H_1^{+}]}} \) \(pH_2 = -\log_{10}{\mathrm{[H_2^{+}]}} \) Since pH decreases with increasing \(\mathrm{[H^+]}\), we can assume that \(pH_2 = pH_1 - 2.00\). Now, \(\mathrm{[H_2^{+}]} = 10^{-(pH_1 - 2)} = 10^{-(pH_1)} \times 10^2 = \mathrm{[H_1^{+}]} \times 10^2\) So, the change in hydrogen ion concentration is a factor of \(10^2\) or 100 times for a pH change of 2.00 units.
05

Step 3(b): Calculate the factor of change for 0.50 units of pH change

Given \(\Delta pH = 0.50\), let's find the hydrogen ion concentration at the two pH values: \(pH_1 = -\log_{10}{\mathrm{[H_1^{+}]}} \) \(pH_2 = -\log_{10}{\mathrm{[H_2^{+}]}} \) Since pH decreases with increasing \(\mathrm{[H^+]}\), we can assume that \(pH_2 = pH_1 - 0.50\). Now, \(\mathrm{[H_2^{+}]} = 10^{-(pH_1 - 0.5)} = 10^{-(pH_1)} \times 10^{0.5} = \mathrm{[H_1^{+}]} \times 10^{0.5}\) So, the change in hydrogen ion concentration is a factor of \(10^{0.5}\) or approximately 3.16 times for a pH change of 0.50 units.

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

Calculate the concentration of an aqueous solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that has a \(\mathrm{pH}\) of \(10.05 .\)

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Explain the following observations: (a) HCl is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S} ;\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is a stronger acid than \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) is a stronger acid than \(\mathrm{HBrO}_{2}\); (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is a stronger acid than \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} ;(\mathrm{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is a stronger acid than phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

(a) Which of the following is the stronger Bronsted-Lowry acid, HBrO or HBr? (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{F}^{-}\) or \(\mathrm{Cl}^{-}\) ? Briefly explain your choices.

The hypochlorite ion, \(\mathrm{ClO}^{-},\) acts as a weak base. (a) Is \(\mathrm{ClO}^{-}\) a stronger or weaker base than hydroxylamine? (b) When ClO \(^{-}\) acts as a base, which atom, Cl or \(\mathrm{O}\), acts as the proton acceptor? (c) Can you use formal charges to rationalize your answer to part (b)?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

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