Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 116

Calculate the number of \(\mathrm{H}^{+}(a q)\) ions in \(1.0 \mathrm{~mL}\) of pure water at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
There are approximately \(6.022 \times 10^{13}\) \(\mathrm{H}^{+}(a q)\) ions in \(1.0 \mathrm{~mL}\) of pure water at \(25^{\circ} \mathrm{C}\).

Step by step solution

01

Identify the ion product constant of water.

At \(25^{\circ} \mathrm{C}\), the ion product constant of water, \(K_{w}\), is given by: \[K_{w} = [\mathrm{H}^{+}][\mathrm{OH}^{-}] = 1.0 \times 10^{-14}\] The concentration of \(\mathrm{H}^{+}(a q)\) and \(\mathrm{OH}^{-}(a q)\) ions is equal in pure water.
02

Calculate the \(\mathrm{H}^{+}\) concentration.

Since the concentration of \(\mathrm{H}^{+}(a q)\) and \(\mathrm{OH}^{-}(a q)\) ions is equal in pure water, we can write the equation \[[\mathrm{H}^{+}]^2 = K_{w}\] Now, calculate the concentration of \(\mathrm{H}^{+}(a q)\) ions: \[[\mathrm{H}^{+}] = \sqrt{K_{w}} = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{mol/L}\]
03

Convert volume to liters.

Since we're given the volume in milliliters, we need to convert it to liters before proceeding. \[1.0 \, \text{mL} = \frac{1.0}{1000} \, \text{L} = 1.0 \times 10^{-3} \, \text{L}\]
04

Calculate the number of \(\mathrm{H}^{+}(a q)\) ions.

To find the number of \(\mathrm{H}^{+}(a q)\) ions in \(1.0 \mathrm{~mL}\) of pure water, we simply multiply the concentration of \(\mathrm{H}^{+}(a q)\) ions with the volume of water in liters: \[\text{Number of } \mathrm{H}^{+}\text{ ions} = [\text{H}^{+}](V) = (1.0 \times 10^{-7} \, \text{mol/L})(1.0 \times 10^{-3}\, \text{L}) = 1.0 \times 10^{-10}\, \text{mol}\]
05

Convert moles to ions.

Finally, we will convert the moles of \(\mathrm{H}^{+}(a q)\) ions to the actual number of ions using Avogadro's number: \[\text{Number of }\mathrm{H}^{+}\text{ ions} = 1.0 \times 10^{-10} \, \text{mol} \times \frac{6.022 \times 10^{23} \, \text{ions}}{1\, \text{mol}}=6.022\times 10^{13}\, \text{ions}\] So, there are approximately \(6.022 \times 10^{13}\) \(\mathrm{H}^{+}(a q)\) ions in \(1.0 \mathrm{~mL}\) of pure water at \(25^{\circ} \mathrm{C}\).

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

If a neutral solution of water, with \(\mathrm{pH}=7.00,\) is cooled to \(10^{\circ} \mathrm{C},\) the pH rises to \(7.27 .\) Which of the following three statements is correct for the cooled water: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\) (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right], \mathrm{or}\) (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in 500.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 10.0 \mathrm{~mL}\) of $0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\( diluted to \)500.0 \mathrm{~mL},(\mathbf{d})\( a solution formed by mixing \)20.0 \mathrm{~mL}$ of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of $8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}$.

At \(50^{\circ} \mathrm{C}\), the ion-product constant for $\mathrm{H}_{2} \mathrm{O}\( has the value \)K_{w}=5.48 \times 10^{-14} \cdot(\mathbf{a})$ What is the \(\mathrm{pH}\) of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of 9.95. Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

Predict the stronger acid in each pair: (a) \(\mathrm{HCl}\) or HF; (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) or \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) or \(\mathrm{HBrO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) or \(\mathrm{HC}_{2} \mathrm{O}_{4} \overline{;} ;(\mathbf{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) or phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right) .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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