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Chapter 14: Problem 43

Calculate the \(\left[\mathrm{OH}^{-}\right]\) of each of the following solutions at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\) c. \(\left[\mathrm{H}^{+}\right]=12 \mathrm{M}\) b. \(\left[\mathrm{H}^{+}\right]=8.3 \times 10^{-16} M\) d. \(\left[\mathrm{H}^{+}\right]=5.4 \times 10^{-5} M\)

Short Answer

Expert verified
a. \([\mathrm{OH}^-] = 1.0 \times 10^{-7} M\) (Neutral) b. \([\mathrm{OH}^-] \approx 1.20 \times 10^{-1} M\) (Basic) c. \([\mathrm{OH}^-] \approx 8.33 \times 10^{-16} M\) (Acidic) d. \([\mathrm{OH}^-] \approx 1.85 \times 10^{-10} M\) (Acidic)

Step by step solution

01

a. Calculate the concentration of OH⁻ ions

Given the concentration of H⁺ ions as \(1.0 \times 10^{-7} M\), we can find the concentration of OH⁻ ions using the Kw value and the formula mentioned above. \(K_w = [\mathrm{H}^+][\mathrm{OH}^-]\) Rearranging for [OH⁻]: \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]}\) Plug in the values: \([\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}}\) \([\mathrm{OH}^-] = 1.0 \times 10^{-7} M\) Since [H⁺] = [OH⁻], the solution is neutral.
02

b. Calculate the concentration of OH⁻ ions

Given the concentration of H⁺ ions as \(8.3 \times 10^{-16} M\), we can find the concentration of OH⁻ ions using the Kw value and the formula mentioned above. \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]}\) Plug in the values: \([\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{8.3 \times 10^{-16}}\) \([\mathrm{OH}^-] \approx 1.20 \times 10^{-1} M\) Since [OH⁻] > [H⁺], the solution is basic.
03

c. Calculate the concentration of OH⁻ ions

Given the concentration of H⁺ ions as \(12 M\), we can find the concentration of OH⁻ ions using the Kw value and the formula mentioned above. \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]}\) Plug in the values: \([\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{12}\) \([\mathrm{OH}^-] \approx 8.33 \times 10^{-16} M\) Since [OH⁻] < [H⁺], the solution is acidic.
04

d. Calculate the concentration of OH⁻ ions

Given the concentration of H⁺ ions as \(5.4 \times 10^{-5} M\), we can find the concentration of OH⁻ ions using the Kw value and the formula mentioned above. \([\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}^+]}\) Plug in the values: \([\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{5.4 \times 10^{-5}}\) \([\mathrm{OH}^-] \approx 1.85 \times 10^{-10} M\) Since [OH⁻] < [H⁺], the solution is acidic.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.050 \mathrm{M} \mathrm{NaCN}\)

What mass of \(\mathrm{KOH}\) is necessary to prepare \(800.0 \mathrm{~mL}\) of a solution having a \(\mathrm{pH}=11.56\) ?

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Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) solution. Assume \(K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\)

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\), determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a \(0.100 M\) solution.
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