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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 \(1.3 脳 10^{-7}\). 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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculation
pH is a crucial measure in chemistry that indicates the acidity or basicity of a solution. It is calculated based on the concentration of hydrogen ions \( ext{H}^+\) in a solution. The formula used is: \[pH = -\log{[\text{H}^+]}\]This calculation can determine how acidic or basic a solution is. A low pH value indicates a high concentration of \( ext{H}^+\), meaning the solution is acidic. Conversely, a high pH value indicates a low concentration of \( ext{H}^+\), which means the solution is basic.
  • For acidic solutions, a pH less than 7 is typical.
  • Neutral solutions, such as pure water, have a pH of exactly 7.
  • Basic, or alkaline, solutions have a pH greater than 7.
In particular, calculating pH involves determining the \( ext{[H}^+]\) concentration using the equilibrium expressions of dissociation reactions. Once the \( ext{[H}^+]\) is known, pH can be directly calculated. Remember, since \(pH\) and \(pOH\) are related by the equation \(pH + pOH = 14\), we can calculate one if the other is known.
Dissociation Reactions
Dissociation reactions are a vital concept in acid-base equilibria as they define how compounds separate into ions in a solution. When salts dissolve in water, they dissociate into their component ions. This can be exemplified by potassium hypobromite \( ext{KBrO}\), which dissociates into potassium ions \( ext{K}^+\) and hypobromite ions \( ext{BrO}^-\).
  • Recognize the types of ions produced: Different salts will dissociate into different positive cations and negative anions.
  • Identify the parent acid or base: For instance, in potassium hypobromite, \( ext{BrO}^-\) is derived from hypobromous acid.
The importance of these reactions lies in their role in determining the composition of solutions. The concentration of ions directly influences the equilibrium of reactions and the ensuing \(pH\)of the solution. To quantify dissociation, we look at equilibrium expressions that describe the behavior of these ions in a solution. Understanding dissociation reactions helps in predicting how a substance will behave in a given environment and influence solution properties.
Equilibrium Expressions
Equilibrium expressions are mathematical representations that express the concentrations of reactants and products at equilibrium. In acid-base chemistry, they help us relate dissociation levels of weak acids or bases to their ion concentrations. These expressions are particularly useful in calculating hydrogen ion concentration \([\text{H}^+]\) and hydroxide ion concentration \([\text{OH}^-]\) of a solution:
  • For a base, such as potassium hypobromite, the equilibrium expression might be \([\text{OH}^-] = K_b[ \text{BrO}^- ]\). A similar approach is used for acids.
The equilibrium constant, either \(K_a\) for acids or \(K_b\) for bases, indicates the tendency of a compound to donate or accept protons.
By understanding equilibrium expressions, one can calculate the extent to which an acid or base will dissociate in water. This is crucial for calculating pH or pOH accurately. Accurately solving for equilibrium conditions involves using known concentrations of compounds and equilibrium constants, which lead to the determination of ion concentrations in solution.
Equilibrium expressions are not only about balance, they showcase the inherent tendencies of molecules and the dynamics of chemical interactions occurring in aqueous environments.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(8.3 \times 10^{-4} \mathrm{MHCl},(\mathbf{b}) 1.20 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(500 \mathrm{~mL}\) of solution, \((\mathbf{c}) 2.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HClO}_{4}\) diluted to \(40.0 \mathrm{~mL}\), (d) a solution formed by mixing \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HBr}\) with \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\).

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6 . What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?

(a) Give the conjugate base of the following Br酶nsted Lowry acids: (i) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) (ii) HBr. (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CN}^{-},\) (ii) \(\mathrm{HSO}_{4}^{-}\).

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that \(\mathrm{pH}\) in the stomach is 2.5 , indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, \(K_{b}=7 \times 10^{-7}\); caffeine, \(K_{b}=4 \times 10^{-14} ;\) strychnine, \(K_{b}=1 \times 10^{-6} ;\) quinine, \(K_{b}=1.1 \times 10^{-6} .\)

Indicate whether each of the following statements is correct or incorrect. (a) Every Br酶nsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.
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