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Chapter 14: Problem 49

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. \(\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\) b. \(\left[\mathrm{H}^{+}\right]=8.3 \times 10^{-16} M\) c. \(\left[\mathrm{H}^{+}\right]=12 M\) d. \(\left[\mathrm{H}^{+}\right]=5.4 \times 10^{-5} M\) Also calculate the pH and pOH of each of these solutions.

Short Answer

Expert verified
a. \(\left[\mathrm{OH}^{-}\right] = 1.0 \times 10^{-7} M\), neutral, pH = 7.0, pOH = 7.0 b. \(\left[\mathrm{OH}^{-}\right] = 1.2 \times 10^{-2} M\), basic, pH = 15.08, pOH = 1.92 c. \(\left[\mathrm{OH}^{-}\right] \approx 8.3 \times 10^{-16} M\), acidic, pH ≈ -1.08, pOH = 15.08 d. \(\left[\mathrm{OH}^{-}\right] \approx 1.9 \times 10^{-10} M\), acidic, pH = 4.27, pOH = 9.73

Step by step solution

01

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}} = 1.0 \times 10^{-7} M\)
02

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] = \left[\mathrm{OH}^{-}\right]\), this solution is neutral.
03

Calculate pH and pOH

pH = \(-\log_{10}(1.0 \times 10^{-7}) = 7.0\) pOH = \(-\log_{10}(1.0 \times 10^{-7}) = 7.0\) b.
04

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{8.3 \times 10^{-16}} = 1.2 \times 10^{-2} M\)
05

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] < \left[\mathrm{OH}^{-}\right]\), this solution is basic.
06

Calculate pH and pOH

pH = \(-\log_{10}(8.3 \times 10^{-16}) = 15.08\) pOH = \(-\log_{10}(1.2 \times 10^{-2}) = 1.92\) c.
07

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{12} \approx 8.3 \times 10^{-16} M\)
08

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), this solution is acidic.
09

Calculate pH and pOH

pH = −\(\log_{10}(12)\) ≈ −1.08 pOH = \(-\log_{10}(8.3 \times 10^{-16}) = 15.08\) d.
10

Calculate OH⁻ Concentration

Using the ion product of water formula: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{w}}{\left[\mathrm{H}^{+}\right]} = \frac{1.0 \times 10^{-14}}{5.4 \times 10^{-5}} \approx 1.9 \times 10^{-10} M\)
11

Identify as Neutral, Acidic, or Basic

Since \(\left[\mathrm{H}^{+}\right] > \left[\mathrm{OH}^{-}\right]\), this solution is acidic.
12

Calculate pH and pOH

pH = \(-\log_{10}(5.4 \times 10^{-5}) = 4.27\) pOH = \(-\log_{10}(1.9 \times 10^{-10}) = 9.73\)

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Most popular questions from this chapter

Determine the pH of a \(0.50-M\) solution of \(\mathrm{NH}_{4} \mathrm{OCl.}\) . See Exercise \(181 .\) )

For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water, $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6} .\) Calculate the concentrations of all species and the pH of a \(2.0-M\) solution of hydrazine in water.

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

When someone hyperventilates, a condition known as respiratory alkalosis can occur. Explain the cause and effect of respiratory alkalosis. Hint: Reference Exercises 146 and 147 .

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated pH. Calculate the pH of each solution. $$\begin{aligned} \text { Group I: } & 0.20 M \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MCl}, 0.20 M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \\ & 0.20 M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl} \end{aligned}$$ $$\begin{aligned} \text { Group II: } 0.20 \quad M\quad \mathrm{KOI}, 0.20 \quad\mathrm{M} \quad\mathrm{NaCN}, 0.20\quad \mathrm{M}\quad \mathrm{KOCl}, 0.20 \\ \mathrm{M}\quad \mathrm{NaNO}_{2} \end{aligned}$$ a. the solution with the lowest pH b. the solution with the highest pH c. the solution with the pH closest to 7.00
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