Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\mathrm{pH}=7.40\) (the normal \(\mathrm{pH}\) of blood) b. \(\mathrm{pII}=15.3\) c. \(\mathrm{pH}=-1.0\) d. \(\mathrm{pH}=3.20\) e. \(\mathrm{pOH}=5.0\) f. \(\mathrm{pOH}=9.60\)

Short Answer

Expert verified
a. \(\left[\mathrm{H}^{+}\right] = 3.98 \times 10^{-8} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 2.51 \times 10^{-7} \mathrm{M}\), basic. b. \(\left[\mathrm{H}^{+}\right] = 5.01 \times 10^{-16} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 2 \times 10^2 \mathrm{M}\), basic. c. \(\left[\mathrm{H}^{+}\right] = 1 \times 10^1 \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 1 \times 10^{-15} \mathrm{M}\), acidic. d. \(\left[\mathrm{H}^{+}\right] = 6.31 \times 10^{-4} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 1.58 \times 10^{-11} \mathrm{M}\), acidic. e. \(\left[\mathrm{H}^{+}\right] = 1 \times 10^{-9} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 1 \times 10^{-5} \mathrm{M}\), basic. f. \(\left[\mathrm{H}^{+}\right] = 2.51 \times 10^{-5} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 3.98 \times 10^{-10} \mathrm{M}\), acidic.

Step by step solution

01

Calculate \(\left[\mathrm{H}^{+}\right]\) for a given pH value

To calculate the concentration of hydrogen ions (\(\left[\mathrm{H}^{+}\right]\)), we use the formula: \[\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\] So, \[\left[\mathrm{H}^{+}\right] = 10^{-7.40} = 3.98 \times 10^{-8} \mathrm{M}\]
02

Calculate pOH and \(\left[\mathrm{OH}^{-}\right]\)

We first calculate pOH using the relationship \(pK_w = pH + pOH\). Therefore, \(pOH = pK_w - pH = 14 - 7.40 = 6.60\). Now, to calculate the concentration of hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)), we use the formula: \[\left[\mathrm{OH}^{-}\right] = 10^{-\mathrm{pOH}}\] So, \[\left[\mathrm{OH}^{-}\right] = 10^{-6.60} = 2.51 \times 10^{-7} \mathrm{M}\]
03

Identify as neutral, acidic, or basic

Since the value of the given pH (7.40) is greater than 7, the solution is basic. b. \(\mathrm{pH}=15.3\) Repeat the steps from solution a. We calculate \(\left[\mathrm{H}^{+}\right]\), pOH, and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the given pH (15.3) is greater than 7, the solution is basic. c. \(\mathrm{pH}=-1.0\) Repeat the steps from solution a. We calculate \(\left[\mathrm{H}^{+}\right]\), pOH and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the given pH (-1.0) is less than 7, the solution is acidic. d. \(\mathrm{pH}=3.20\) Repeat the steps from solution a. We calculate \(\left[\mathrm{H}^{+}\right]\), pOH and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the given pH (3.20) is less than 7, the solution is acidic. e. \(\mathrm{pOH}=5.0\)
04

Calculate pH and \(\left[\mathrm{H}^{+}\right]\)

We first calculate pH using the relationship \(pK_w = pH + pOH\). Therefore, \(pH = pK_w - pOH = 14 - 5.0 = 9.0\). Now, to calculate the concentration of hydrogen ions (\(\left[\mathrm{H}^{+}\right]\)), we use the formula: \[\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\] So, \[\left[\mathrm{H}^{+}\right] = 10^{-9.0} = 1 \times 10^{-9} \mathrm{M}\]
05

Calculate \(\left[\mathrm{OH}^{-}\right]\)

To calculate the concentration of hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)), we use the formula: \[\left[\mathrm{OH}^{-}\right] = 10^{-\mathrm{pOH}}\] So, \[\left[\mathrm{OH}^{-}\right] = 10^{-5.0} = 1 \times 10^{-5} \mathrm{M}\]
06

Identify as neutral, acidic, or basic

Since the value of the calculated pH (9.0) is greater than 7, the solution is basic. f. \(\mathrm{pOH}=9.60\) Repeat the steps from solution e. We calculate pH, \(\left[\mathrm{H}^{+}\right]\), and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the calculated pH is less than 7, the solution is acidic.

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!

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

At \(25^{\circ} \mathrm{C}\), a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times\right.\) \(10^{-5}\) ) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}{ }^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.)

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but is now available only by prescription because of its addictive properties. If the \(\mathrm{pH}\) of a \(1.7 \times 10^{-3}-M\) solution of codeine is \(9.59\), calculate \(K_{\mathrm{b}}\).

The \(\mathrm{pH}\) of a \(0.063-M\) solution of hypobromous acid \((\mathrm{HOBr}\) but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

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