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Chapter 13: Problem 15

Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in solutions with the following \(\mathrm{pH}\). (a) \(4.0\) (b) \(8.52\) (c) \(0.00\) (d) \(12.60\)

Short Answer

Expert verified
Given the pH values, calculate the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in the following solutions: a) pH = 4.0 [H+] = \(1.0 \times 10^{-4}\) M, [OH-] = \(1.0 \times 10^{-10}\) M. b) pH = 8.52 [H+] = \(2.91 \times 10^{-9}\) M, [OH-] = \(3.31 \times 10^{-6}\) M. c) pH = 0.00 [H+] = \(1.0\) M, [OH-] = \(1.0 \times 10^{-14}\) M. d) pH = 12.60 [H+] = \(2.51 \times 10^{-13}\) M, [OH-] = \(3.98 \times 10^{-2}\) M.

Step by step solution

01

Calculate [H+] using pH

Use the pH equation to find the concentration of hydrogen ions (H+): pH = -log[H+] => [H+] = \(10^{-\mathrm{pH}}\) = \(10^{-4.0}\) = \(1.0 \times 10^{-4}\) M.
02

Calculate pOH

Since pH + pOH = 14, we can find the pOH: pOH = 14 - pH = 14 - 4.0 = 10.0
03

Calculate [OH-] using pOH

Use the pOH equation to find the concentration of hydroxide ions (OH-): pOH = -log[OH-] => [OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-10.0}\) = \(1.0 \times 10^{-10}\) M. #a) Concentrations: [H+] = \(1.0 \times 10^{-4}\) M, [OH-] = \(1.0 \times 10^{-10}\) M. #b) 8.52#
04

Calculate [H+] using pH

Use the pH equation to find the concentration of hydrogen ions (H+): [H+] = \(10^{-\mathrm{pH}}\) = \(10^{-8.52}\) = \(2.91 \times 10^{-9}\) M.
05

Calculate pOH

Since pH + pOH = 14, we can find the pOH: pOH = 14 - pH = 14 - 8.52 = 5.48
06

Calculate [OH-] using pOH

Use the pOH equation to find the concentration of hydroxide ions (OH-): [OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-5.48}\) = \(3.31 \times 10^{-6}\) M. #b) Concentrations: [H+] = \(2.91 \times 10^{-9}\) M, [OH-] = \(3.31 \times 10^{-6}\) M. #c) 0.00#
07

Calculate [H+] using pH

Use the pH equation to find the concentration of hydrogen ions (H+): [H+] = \(10^{-\mathrm{pH}}\) = \(10^{-0.0}\) = \(1.0\) M.
08

Calculate pOH

Since pH + pOH = 14, we can find the pOH: pOH = 14 - pH = 14 - 0.0 = 14.0
09

Calculate [OH-] using pOH

Use the pOH equation to find the concentration of hydroxide ions (OH-): [OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-14.0}\) = \(1.0 \times 10^{-14}\) M. #c) Concentrations: [H+] = \(1.0\) M, [OH-] = \(1.0 \times 10^{-14}\) M. #d) 12.60#
10

Calculate [H+] using pH

Use the pH equation to find the concentration of hydrogen ions (H+): [H+] = \(10^{-\mathrm{pH}}\) = \(10^{-12.60}\) = \(2.51 \times 10^{-13}\) M.
11

Calculate pOH

Since pH + pOH = 14, we can find the pOH: pOH = 14 - pH = 14 - 12.60 = 1.40
12

Calculate [OH-] using pOH

Use the pOH equation to find the concentration of hydroxide ions (OH-): [OH-] = \(10^{-\mathrm{pOH}}\) = \(10^{-1.40}\) = \(3.98 \times 10^{-2}\) M. #d) Concentrations: [H+] = \(2.51 \times 10^{-13}\) M, [OH-] = \(3.98 \times 10^{-2}\) M.

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Most popular questions from this chapter

Caproic acid, \(\mathrm{HC}_{6} \mathrm{H}_{11} \mathrm{O}_{2}\), is found in coconut oil and is used in making artificial flavors. A solution is made by dissolving \(0.450 \mathrm{~mol}\) of caproic acid in enough water to make \(2.0 \mathrm{~L}\) of solution. The solution has \(\left[\mathrm{H}^{+}\right]=1.7 \times\) \(10^{-3} M\). What is \(K_{\mathrm{a}}\) for caproic acid?

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.

Write the ionization equation and the \(K_{\mathrm{a}}\) expression for each of the following acids. (a) \(\mathrm{PH}_{4}{ }^{+}\) (b) \(\mathrm{HS}^{-}\) (c) \(\mathrm{HBrO}_{2}\)

The solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) at \(25^{\circ} \mathrm{C}\) is \(0.153 \mathrm{~g} / 100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). Assuming that the density of a saturated solution is \(1.00 \mathrm{~g} / \mathrm{mL}\), calculate the maximum \(\mathrm{pH}\) one can obtain when \(\mathrm{Ca}(\mathrm{OH})_{2}\) is dissolved in water.

Using the Tables in Appendix 1, calculate \(\Delta H\) for the reaction of the following. (a) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) (b) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M}\) HF, taking the heat of formation of \(\mathrm{HF}(a q)\) to be \(-320.1 \mathrm{~kJ} / \mathrm{mol}\)
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