Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 24

Milk of Magnesia has a pH of \(10.5\). (a) Calculate \(\left[\mathrm{H}^{+}\right]\). (b) Calculate the ratio of the \(\mathrm{H}^{+}\) concentration of gastric juice, \(\mathrm{pH} 1.5\), to that of Milk of Magnesia.

Short Answer

Expert verified
Answer: The concentration of H+ ions in Milk of Magnesia is 3.16 × 10^{-11} M, and the concentration in gastric juice is 3.16 × 10^{-2} M. The ratio of the concentration of H+ ions of gastric juice to that of Milk of Magnesia is 1,000,000,000:1 or 10^9:1.

Step by step solution

01

Calculate the concentration of H+ ions in Milk of Magnesia

Since pH = -log[H+], we can rewrite the formula to find the concentration of H+ ions ( [H+] ): [H+] = 10^(-pH) The question gives the pH of Milk of Magnesia as 10.5, so we can plug this value into the formula: [H+] = 10^(-10.5) Now, calculating the concentration of H+ ions: [H+] = 3.16 × 10^{-11} M
02

Calculate the concentration of H+ ions in gastric juice

Repeat the same process for gastric juice, with a given pH value of 1.5: [H+] = 10^(-1.5) Now, calculating the concentration of H+ ions in gastric juice: [H+] = 3.16 × 10^{-2} M
03

Calculate the ratio of H+ concentration of gastric juice to H+ concentration of Milk of Magnesia

To calculate the ratio, divide the concentration of H+ ions in gastric juice by the concentration of H+ ions in Milk of Magnesia: Ratio = (3.16 × 10^{-2} M) / (3.16 × 10^{-11} M) The units of Molarity (M) will cancel out: Ratio = 10^9 So, the ratio of the concentration of H+ ions of gastric juice to that of Milk of Magnesia is 1,000,000,000:1 or 10^9:1.

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

Using the Brønsted-Lowry model, write an equation to show why each of the following species produces a basic aqueous solution. (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{NO}_{2}^{-}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) \(\mathrm{CO}_{3}{ }^{2-}\) (e) \(\mathrm{F}^{-}\) (f) \(\mathrm{HCO}_{3}^{-}\)

WEB Give the formula of the conjugate base of (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)(\mathrm{OH})_{3}^{-}\) (c) \(\mathrm{HNO}_{2}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}\) (e) \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

What is the \(\mathrm{pH}\) of a solution obtained by adding \(5.00 \mathrm{~g}\) of \(\mathrm{HI}\) to \(295 \mathrm{~mL}\) of a \(0.786 M\) solution of \(\mathrm{HNO}_{3} ?\) Assume that the HI addition does not change the volume of the resulting solution.

Write the ionization equation and the \(K_{\mathrm{a}}\) expression for each of the following acids. (a) \(\mathrm{HSO}_{3}^{-}\) (b) \(\mathrm{HPO}_{4}{ }^{2-}\) (c) \(\mathrm{HNO}_{2}\)

Write the ionization expression and the \(K_{b}\) expression for \(0.1 M\) aqueous solutions of the following bases. (a) \(\mathrm{F}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}\) (c) \(\mathrm{CN}^{-}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic 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