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Chapter 16: Problem 79

Using data from Appendix \(D\), calculate \(p O H\) and \(p H\) for each (a) \(0.080 M\) potassium hypobromite of the following solutions: \((\mathrm{KBrO}),\) (b) \(0.150 \mathrm{M}\) potassium hydrosulfide \((\mathrm{KHS}),(\mathbf{c})\) a mixture that is \(0.25 \mathrm{M}\) in potassium nitrite \(\left(\mathrm{KNO}_{2}\right)\) and \(0.15 \mathrm{M}\) in magnesium nitrite \(\left(\mathrm{Mg}\left(\mathrm{NO}_{2}\right)_{2}\right)\).

Short Answer

Expert verified
For 0.080 M potassium hypobromite, the \(pOH = 1.097\) and the \(pH = 12.903\). For a 0.150 M potassium hydrosulfide solution, the \(pOH = 10.675\) and the \(pH = 3.325\). For the mixture with \(0.25 \, \mathrm{M}\) potassium nitrite and \(0.15 \, \mathrm{M}\) magnesium nitrite, the \(pOH = 9.18\) and the \(pH = 4.82\).

Step by step solution

01

Identify the dissociation reaction

: The dissociation reaction of potassium hypobromite is: \[\mathrm{KBrO} \rightarrow \mathrm{K^+} + \mathrm{BrO^{-}}\]
02

Write the equilibrium expression

: The equilibrium expression for the reaction is: \[\mathrm{[OH^-]} = \mathrm{[BrO^-]}\]
03

Calculate the \(OH^-\) concentration

: Given the concentration of \(\mathrm{KBrO}\), \(\mathrm{[OH^-]} = 0.080 \, \mathrm{M}\)
04

Calculate \(pOH\) and \(pH\)

: \(pOH = -\log\left[\mathrm{OH^-}\right] = -\log\left(0.080\right)\) \(pOH = 1.097\) To find the pH, use the relationship: \(pH + pOH = 14\) \(pH = 14 - pOH = 14 - 1.097 = 12.903\) For 0.080 M potassium hypobromite, the \(pOH = 1.097\) and the \(pH = 12.903\). (b) For \(\mathrm{KHS}\) (potassium hydrosulfide) solution:
05

Identify the dissociation reaction

: The dissociation reaction of potassium hydrosulfide is: \[\mathrm{KHS} \rightarrow \mathrm{K^+} + \mathrm{HS^{-}}\]
06

Write the equilibrium expression

: The equilibrium expression for the reaction is: \[\mathrm{[H^+]} = \mathrm{K_a}\times\frac{\mathrm{[HS^-]}}{\mathrm{[H^+]}}\]
07

Calculate the \(H^+\) concentration

: Given the concentration of \(\mathrm{KHS}\), \(\mathrm{[HS^-]} = 0.150 \, \mathrm{M}\). For \(\mathrm{HS^{-}}\), the acid dissociation constant \(\mathrm{K_a}\) is \(1.3 × 10^{-7}\). Using the equilibrium expression, we can solve for the \(H^+\) concentration: \[\mathrm{[H^+]} = \sqrt{\mathrm{K_a} \times [\mathrm{HS^-}]} = \sqrt{(1.3 \times 10^{-7}) \times 0.150}\] \(\mathrm{[H^+]} = 4.74 \times 10^{-4} \, \mathrm{M}\)
08

Calculate \(pOH\) and \(pH\)

: \(pH = -\log\left[\mathrm{H^+}\right] = -\log\left(4.74 \times 10^{-4}\right)\) \(pH = 3.325\) To find the pOH, use the relationship: \(pH + pOH = 14\) \(pOH = 14 - pH = 14 - 3.325 = 10.675\) For a 0.150 M potassium hydrosulfide solution, the \(pOH = 10.675\) and the \(pH = 3.325\). (c) For \(\mathrm{KNO}_2\) and \(\mathrm{Mg(NO}_2)_2\) mixture: Since the given mixture contains more than one compound, we need to consider both compounds to calculate the \(OH^-\) concentration. The dissociation reactions are: \[\mathrm{KNO_2} \rightarrow \mathrm{K^+} + \mathrm{NO_2^{-}}\] \[\mathrm{Mg(NO_2)_2} \rightarrow \mathrm{Mg^{2+}} + 2\mathrm{NO_2^{-}}\]
09

Write the equilibrium expression

: The equilibrium expression for the reaction is: \[\mathrm{[OH^-]} = \mathrm{K_b[ NO_2^-]}\]
10

Calculate the \(OH^-\) concentration

: Given the concentration of \(\mathrm{KNO}_2\), \(\mathrm{[NO_2^-]} = 0.25 \, \mathrm{M}\) and for \(\mathrm{Mg(NO_2)_2}\), \(\mathrm{[NO_2^-]} = 0.3 \, \mathrm{M}\). The total \(\mathrm{NO_2^{-}}\) concentration is \(0.25 + 0.3 = 0.55 \, \mathrm{M}\). Using the equilibrium expression, we can solve for the \(OH^-\) concentration: \[\mathrm{[OH^-]} = \mathrm{K_b[ NO_2^-]} = 1.2 \times 10^{-9} \times 0.55\] \(\mathrm{[OH^-]} = 6.6 \times 10^{-10} \, \mathrm{M}\)
11

Calculate \(pOH\) and \(pH\)

: \(pOH = -\log\left[\mathrm{OH^-}\right] = -\log\left(6.6 \times 10^{-10}\right)\) \(pOH = 9.18\) To find the pH, use the relationship: \(pH + pOH = 14\) \(pH = 14 - pOH = 14 - 9.18 = 4.82\) For the mixture with \(0.25 \, \mathrm{M}\) potassium nitrite and \(0.15 \, \mathrm{M}\) magnesium nitrite, the \(pOH = 9.18\) and the \(pH = 4.82\).

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Most popular questions from this chapter

At \(50^{\circ} \mathrm{C}\), the ion-product constant for $\mathrm{H}_{2} \mathrm{O}\( has the value \)K_{w}=5.48 \times 10^{-14} \cdot(\mathbf{a})$ What is the \(\mathrm{pH}\) of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

If a neutral solution of water, with \(\mathrm{pH}=7.00,\) is cooled to \(10^{\circ} \mathrm{C},\) the pH rises to \(7.27 .\) Which of the following three statements is correct for the cooled water: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\) (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right], \mathrm{or}\) (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. (i) $\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ (ii) $\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)$ (b) We usually use single arrows for reactions when the forward reaction is appreciable ( \(K\) much greater than 1) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in $200 \mathrm{~mL}$ of water?

Predict the stronger acid in each pair: (a) \(\mathrm{HCl}\) or HF; (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) or \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) or \(\mathrm{HBrO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) or \(\mathrm{HC}_{2} \mathrm{O}_{4} \overline{;} ;(\mathbf{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) or phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right) .\)
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