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Chapter 16: Problem 137

A generic base, \(\mathrm{B}^{-}\), is added to \(2.25 \mathrm{~L}\) of water. The \(\mathrm{pH}\) of the solution is found to be \(10.10\). What is the concentration of the base \(\mathrm{B}^{-}\) in this solution? \(K_{a}\) for the acid \(\mathrm{HB}\) at \(25^{\circ} \mathrm{C}\) is \(1.99 \times 10^{-9}\).

Short Answer

Expert verified
The concentration of the base \(\text{B}^-\) is approximately \(3.15 \times 10^{-3}\) M.

Step by step solution

01

Calculate the pOH

The pH of the solution is given as 10.10. To find the pOH, use the relation: \( \text{pH} + \text{pOH} = 14 \). Therefore, \( \text{pOH} = 14 - 10.10 = 3.90 \).
02

Calculate Hydroxide Ion Concentration

The pOH of the solution is 3.90. To find the hydroxide ion concentration \([\text{OH}^-]\), use the formula: \( [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-3.90} \approx 1.26 \times 10^{-4} \text{ M} \).
03

Use the Base Dissociation Constant

Given that \( K_a = 1.99 \times 10^{-9} \), we can find the base dissociation constant \( K_b \) using the relation \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \). Thus, \( K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.99 \times 10^{-9}} \approx 5.03 \times 10^{-6} \).
04

Determine the Base Concentration

For a weak base \( B^- \), the equilibrium relation is \( [\text{OH}^-] = \sqrt{K_b \times [B^-]} \). Solving for \([B^-]\), we have: \( [B^-] = \frac{[\text{OH}^-]^2}{K_b} = \frac{(1.26 \times 10^{-4})^2}{5.03 \times 10^{-6}} \approx 3.15 \times 10^{-3} \text{ M} \).

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Key Concepts

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

pH Calculations
Understanding pH is crucial when working with solutions in chemistry. pH is a measure of the acidity or basicity of an aqueous solution. It is determined by the concentration of hydrogen ions (\([\text{H}^+]\)) present.
For pH calculations, a key aspect to remember is the relationship:
  • \( \text{pH} + \text{pOH} = 14 \)
This equation holds true for aqueous solutions at 25掳C. In this exercise, the pH given was 10.10, which indicates a basic solution.
By solving for the pOH, we calculated it by subtracting the pH from 14. This gives us insight into the concentration of hydroxide ions in the solution, as pOH is a measure of alkalinity.
Hydroxide Ion Concentration
The concentration of hydroxide ions \([\text{OH}^-]\) in a solution is directly related to whether the solution behaves as acidic or basic. Once pOH is known, we can calculate \([\text{OH}^-]\) using the formula:
  • \( [\text{OH}^-] = 10^{-\text{pOH}} \)
Here, the pOH was 3.90, and substituting into the formula gave us the hydroxide ion concentration as roughly \(1.26 \times 10^{-4}\) M.
Remember, the lower the pOH, the higher the concentration of \([\text{OH}^-]\), indicating a stronger basic character.
Accurate knowledge of hydroxide ion concentration is essential for understanding the solution's behavior and for further calculations involving equilibrium constants.
Weak Base Equilibrium
Weak base equilibrium involves understanding how a weak base interacts with water to form an equilibrium between the base, its conjugate acid, and hydroxide ions. For a weak base \(B^-\), the equilibrium relation follows:
  • \( [\text{OH}^-] = \sqrt{K_b \times [B^-]} \)
To find the concentration of the base \([B^-]\), we rearrange to:
  • \( [B^-] = \frac{[\text{OH}^-]^2}{K_b} \)
In this problem, calculations depend on knowing \(K_b\), the base dissociation constant derived from \(K_a\) for the conjugate acid using \( K_w = K_a \times K_b \).
This approach involves leveraging the interrelation between the acid and base equilibrium constants, allowing us to determine unknown concentrations in solution. The equilibrium analysis described provides a framework for calculating different species in solution, essential for understanding the behavior of weak bases.

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Most popular questions from this chapter

\(K_{a}\) for formic acid is \(1.7 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). A buffer is made by mixing \(529 \mathrm{~mL}\) of \(0.465 \mathrm{M}\) formic acid, \(\mathrm{HCHO}_{2}\), and \(494 \mathrm{~mL}\) of \(0.524 M\) sodium formate, \(\mathrm{NaCHO}_{2}\). Calculate the \(\mathrm{pH}\) of this solution at \(25^{\circ} \mathrm{C}\) after \(110 . \mathrm{mL}\) of \(0.152 \mathrm{M} \mathrm{HCl}\) has been added to this buffer.

Write the equation for the acid ionization of the \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion.

The equilibrium equations and \(K_{a}\) values for three reaction systems are given below. \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{NH}_{3}(a q) ; K_{a}=5.6 \times 10^{-10}\) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) ; K_{a}=4.3 \times 10^{-7}\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}\) $$ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HPO}_{4}^{2-}(a q) ; K_{a}=6.2 \times 10^{-8} $$ a. Which conjugate pair would be best for preparing a buffer with a pH of \(6.96\) ? Why? b. How would you prepare \(100 \mathrm{~mL}\) of a buffer with a \(\mathrm{pH}\) of \(6.96\) assuming that you had available \(0.10 \mathrm{M}\) solutions of each pair?

A 0.239-g sample of unknown organic base is dissolved in water and titrated with a \(0.135 M\) hydrochloric acid solution. After the addition of \(18.35 \mathrm{~mL}\) of acid, a pH of \(10.73\) is recorded. The equivalence point is reached when a total of \(39.24 \mathrm{~mL}\) of \(\mathrm{HCl}\) is added. The base and acid combine in a 1:1 ratio. a. What is the molar mass of the organic base? b. What is the \(K_{b}\) value for the base? The \(K_{b}\) value could have been determined very easily if a pH measurement had been made after the addition of \(19.62 \mathrm{~mL}\) of HCl. Why?

Calculate the \(\mathrm{pH}\) of a solution obtained by mixing 456 \(\mathrm{mL}\) of \(0.10 M\) hydrochloric acid with \(285 \mathrm{~mL}\) of \(0.15 M\) sodium hydroxide. Assume the combined volume is the sum of the two original volumes.
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