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Chapter 14: Problem 118

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 \mathrm{M} \mathrm{KNO}_{2}\) c. \(0.40 \mathrm{M} \mathrm{NH}_{4} \mathrm{ClO}_{4}\) b. \(0.45 \mathrm{M} \mathrm{NaOCl}\)

Short Answer

Expert verified
The pH of the given solutions are: a. Solution with $0.12 \mathrm{M} \mathrm{KNO}_{2}$: pH $\approx 8.22$ c. Solution with $0.40 \mathrm{M} \mathrm{NH}_{4} \mathrm{ClO}_{4}$: pH $\approx 5.22$ b. Solution with $0.45 \mathrm{M} \mathrm{NaOCl}$: pH $\approx 7.52$

Step by step solution

01

Calculate the concentration of the conjugate acid/base

For each of the solutions, we will first determine the concentration of their respective conjugate acid/base, required for the calculation of pH. For Solution a: \(\mathrm{KNO}_{2}\) dissociates in water to form \(\mathrm{OH}^{-}\) ions and \(\mathrm{NO}_{2}^{-}\) ions. Since the concentration of \(\mathrm{KNO}_{2}\) is given, the concentration of \(\mathrm{NO}_{2}^{-}\) ions will also be \(0.12\)M. For Solution c: \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) dissociates in water to form \(\mathrm{NH}_{4}^{+}\) ions and \(\mathrm{ClO}_{4}^{-}\) ions. The concentration of \(\mathrm{NH}_{4}^{+}\) ions will be \(0.40\)M. For Solution b: \(\mathrm{NaOCl}\) dissociates in water to form \(\mathrm{Na}^{+}\) ions and \(\mathrm{OCl}^{-}\) ions. The concentration of \(\mathrm{OCl}^{-}\) ions will be \(0.45\)M.
02

Determine the \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) concentration

We will now find the concentration of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions in each solution using the equilibrium constants (Ka or Kb) of their respective conjugate acid/base. For Solution a: The equilibrium constant for the dissociation of \(\mathrm{NO}_{2}^{-}\) ions, Kb, is given by: \[K_b = \frac{[\mathrm{HNO}_{2+}][\mathrm{OH}^鈭抅}{[\mathrm{NO}_{2}^{鈭拀]}\] The \(\mathrm{OH}^{-}\) concentration can be calculated as: \[\mathrm{OH}^{-} = \sqrt{K_b \times [\mathrm{NO}_{2}^{-}]}\] For Solution c: The equilibrium constant for the dissociation of \(\mathrm{NH}_{4}^{+}\) ions, Ka, is given by: \[K_a = \frac{[\mathrm{NH}_{3}][\mathrm{H}^+]}{[\mathrm{NH}_{4}^{+}]}\] The \(\mathrm{H}^{+}\) concentration can be calculated as: \[\mathrm{H}^{+} = \sqrt{K_a \times [\mathrm{NH}_{4}^{+}]}\] For Solution b: The equilibrium constant for the dissociation of \(\mathrm{OCl}^{-}\) ions, Kb, is given by: \[K_b = \frac{[\mathrm{HOCl+}][\mathrm{OH}^-]}{[\mathrm{OCl}^{-}]}\] The \(\mathrm{OH}^{-}\) concentration can be calculated as: \[\mathrm{OH}^{-} = \sqrt{K_b \times [\mathrm{OCl}^{-}]}\]
03

Calculate the pH

Finally, we will calculate the pH for each solution using the \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) concentrations. For Solution a: Calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \[pOH = -\log([\mathrm{OH}^{-}])\] Then, calculate pH using the relationship: \[pH = 14 - pOH\] For Solution c: Calculate pH using the \(\mathrm{H}^{+}\) concentration: \[pH = -\log([\mathrm{H}^+])\] For Solution b: Calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \[pOH = -\log([\mathrm{OH}^{-}])\] Then, calculate pH using the relationship: \[pH = 14 - pOH\]

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Key Concepts

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

acid-base equilibrium
Acid-base equilibrium is a key concept in the field of chemistry that describes the balance between acid and base forms of a compound in solution. An acid-base equilibrium is established when a weak acid partially dissociates in water to donate protons (\(\mathrm{H}^+\)), forming its conjugate base. Similarly, a weak base can accept protons, forming its conjugate acid. The equilibrium state is dynamic, meaning reactions proceed in both directions at equal rates:
  • The forward reaction involves the dissociation of the acid into its conjugate base and protons.
  • The reverse reaction involves the recombination of the conjugate base and protons to form the acid again.
This balance is crucial for calculating the pH of solutions, as the concentrations of \(\mathrm{H}^+\) and \(\mathrm{OH}^-\) ions directly affect pH values. Understanding this equilibrium helps students solve problems like the pH calculations in various solutions such as \(\mathrm{KNO}_2\), \(\mathrm{NH}_4\mathrm{ClO}_4\), and \(\mathrm{NaOCl}\) shown in the exercise.
conjugate acid-base pairs
Conjugate acid-base pairs are fundamental to understanding acid-base reactions. When an acid donates a proton, the remaining part is called its conjugate base, while when a base accepts a proton, it becomes a conjugate acid.Each conjugate pair is related through a simple equilibrium:
  • The acid in the pair is capable of donating a proton.
  • The base is capable of accepting a proton.
This relationship is crucial for predicting the behavior of species in solution. For example, in the exercise solutions:
  • \(\mathrm{NO}_2^-\) is the conjugate base of \(\mathrm{HNO}_2\), contributing \(\mathrm{OH}^-\) ions in solution when \(\mathrm{KNO}_2\) is dissolved.
  • \(\mathrm{NH}_4^+\) is the conjugate acid of \(\mathrm{NH}_3\), releasing \(\mathrm{H}^+\) ions when \(\mathrm{NH}_4\mathrm{ClO}_4\) is in solution.
  • \(\mathrm{OCl}^-\) is the conjugate base of \(\mathrm{HOCl}\), affecting \(\mathrm{pH}\) through its dissociation in water.
These pairs help us understand the shifts in pH as well as the reaction paths that balance the acid and base forms in equilibrium.
equilibrium constants
Equilibrium constants are pivotal in calculating the extent of reactions in solutions, particularly acid-base reactions. Defined as \(K_a\) for acids and \(K_b\) for bases, these constants help determine the concentration of \(\mathrm{H}^+\) or \(\mathrm{OH}^-\) ions at equilibrium.The formulas for these constants are as follows:
  • For acid dissociation:\(K_a = \frac{[\mathrm{H}^+][\text{Conjugate Base}]}{[\text{Acid}]}\)
  • For base dissociation:\(K_b = \frac{[\text{Conjugate Acid}][\mathrm{OH}^-]}{[\text{Base}]}\)
Equilibrium constants indicate the strength of acids and bases; the larger the \(K_a\) or \(K_b\) value, the stronger the acid or base. In the exercise, these constants are used to calculate the concentration of ions in solutions of \(\mathrm{KNO}_2\), \(\mathrm{NH}_4\mathrm{ClO}_4\), and \(\mathrm{NaOCl}\), which are essential steps in determining \(\mathrm{pH}\).
solution concentration
Solution concentration is a measure of the amount of solute present in a given volume of solvent. In acid-base chemistry, understanding the concentration of ions, such as \(\mathrm{H}^+\) and \(\mathrm{OH}^-\), is crucial for calculating \(\mathrm{pH}\).Concentration determines the degree to which the acid or base can affect the pH of a solution. Here are some key points:
  • Concentration is usually expressed in molarity (\(\mathrm{M}\)), which is the number of moles of solute per liter of solution.
  • Knowing the initial concentration of solutes like \(\mathrm{KNO}_2\), \(\mathrm{NH}_4\mathrm{ClO}_4\), and \(\mathrm{NaOCl}\) is important for starting calculations.
  • Equilibrium concentrations are used in conjunction with \(K_a\) or \(K_b\) to find the concentrations of ions like \(\mathrm{H}^+\) and \(\mathrm{OH}^-\).
Ultimately, the concentration of the solution influences the resulting \(\mathrm{pH}\), making it one of the foundational aspects to understand in chemistry.

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Most popular questions from this chapter

A solution is prepared by adding \(50.0 \mathrm{~mL}\) of \(0.050 M \mathrm{HBr}\) to \(150.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HI. Calculate the concentrations of all species in this solution. HBr and HI are both considered strong acids.

Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Br酶nsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

Calculate the concentration of an aqueous HI solution that has \(\mathrm{pH}=2.50 .\) HI is a strong acid.

Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\mathrm{pH}=7.40\) (the normal \(\mathrm{pH}\) of blood) b. \(\mathrm{pH}=15.3\) c. \(\mathrm{pH}=-1.0\) d. \(\mathrm{pH}=3.20\) e. \(\mathrm{pOH}=5.0\) f. \(\mathrm{pOH}=9.60\)

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasantly tart taste. Although phosphoric acid is a triprotic acid, its protons are lost one at a time. Assuming that in cola drinks the concentration of phosphoric acid is \(0.007 M\), calculate the \(\mathrm{pH}\) in this solution.
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