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Chapter 14: Problem 100

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

Short Answer

Expert verified
The pH of the 0.050 M diethylamine solution is approximately 11.51.

Step by step solution

01

Write the base ionization reaction

First, we need to write the reaction of diethylamine with water, where it acts as a base by accepting a proton (H+) from water and forms its conjugate acid and hydroxide ions: \[(C_2H_5)_2NH + H_2O \rightleftharpoons (C_2H_5)_2NH_2^{+} + OH^-\]
02

Write the expression for Kb

Now, we need to write the equilibrium expression for the reaction using the provided \(K_b\): \[K_b = \frac{[(C_{2}H_{5})_{2}NH_2^{+}][OH^{-}]}{[(C_{2}H_{5})_{2}NH]}\]
03

Set up the ICE table

Next, we need to set up the Initial, Change, and Equilibrium (ICE) table to keep track of the changes in concentration for each species. \[ \begin{array}{|c|c|c|c|} \hline & (C_2H_5)_2NH & (C_2H_5)_2NH_2^+ & OH^- \\ \hline \text{Initial} & 0.050 & 0 & 0 \\ \hline \text{Change} & -x & +x & +x \\ \hline \text{Equilibrium} & 0.050-x & x & x \\ \hline \end{array} \] Here, "x" represents the change in concentration, which is unknown.
04

Substitute equilibrium concentrations into Kb expression

Substitute the equilibrium concentrations from the ICE table into the \(K_b\) expression and solve for "x": \[1.3 \times 10^{-3} = \frac{x^2}{0.050 - x}\]
05

Solve for x

Since \(K_b\) is relatively small, we can assume that the change in concentration "x" is much smaller than 0.050, so we can simplify the equation as follows: \[1.3 \times 10^{-3} = \frac{x^2}{0.050}\] Now, solve for x: \[x = \sqrt{1.3 \times 10^{-3} \times 0.050} \approx 0.0032\] The value of x represents the equilibrium concentration of hydroxide ions, \([OH^-]\).
06

Calculate the pOH and pH

First, calculate the pOH by taking the negative logarithm of the hydroxide ion concentration: \[pOH = -\log{[OH^-]} = -\log{(0.0032)} \approx 2.49\] Now, we can use the relationship between pH and pOH to find the pH: \[pH + pOH = 14\] \[pH = 14 - pOH = 14 - 2.49 \approx 11.51\] So, the pH of the 0.050 M diethylamine solution is approximately 11.51.

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Most popular questions from this chapter

Consider \(0.10 M\) solutions of the following compounds: \(\mathrm{AlCl}_{3}\) \(\mathrm{NaCN}, \mathrm{KOH}, \mathrm{CsClO}_{4}\), and NaF. Place these solutions in order of increasing \(\mathrm{pH}\).

Calculate the \(\mathrm{pH}\) of an aqueous solution containing \(1.0 \mathrm{X}\) \(10^{-2} M \mathrm{HCl}, 1.0 \times 10^{-2} M \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(1.0 \times 10^{-2} M \mathrm{HCN}\).

Calculate the percent dissociation of the acid in each of the following solutions. a. \(0.50 M\) acetic acid b. \(0.050 M\) acetic acid c. \(0.0050 M\) acetic acid d. Use Le Châtelier's principle to explain why percent dissociation increases as the concentration of a weak acid decreases. e. Even though the percent dissociation increases from solutions a to \(\mathrm{c}\), the \(\left[\mathrm{H}^{+}\right]\) decreases. Explain.

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)
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