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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

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 \(-320.1 \mathrm{~kJ} / \mathrm{mol}\)

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) \(\mathrm{HCO}_{3}^{-}\)

Barbituric acid \(\left(K_{\mathrm{a}}=1.1 \times 10^{-4}\right)\) is used in the manufacture of some sedatives. For a \(0.673 \mathrm{M}\) solution of barbituric acid, calculate (a) \(\left[\mathrm{H}^{+}\right]\) (b) \(\left[\mathrm{OH}^{-}\right]\) (c) \(\mathrm{pH}\) (d) \% ionization

Consider sodium acrylate, \(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2} . K_{\mathrm{a}}\) for acrylic acid (its conjugate acid) is \(5.5 \times 10^{-5}\). (a) Write a balanced net ionic equation for the reaction that makes aqueous solutions of sodium acrylate basic. (b) Calculate \(K_{b}\) for the reaction in (a). (c) Find the \(\mathrm{pH}\) of a solution prepared by dissolving \(1.61 \mathrm{~g}\) of \(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\) in enough water to make \(835 \mathrm{~mL}\) of solution.

Find the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of solutions with the following \(\left[\mathrm{H}^{+}\right]\). Classify each as acidic or basic. (a) \(6.0 \mathrm{M}\) (b) \(0.33 \mathrm{M}\) (c) \(4.6 \times 10^{-8} M\) (d) \(7.2 \times 10^{-14} M\)
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