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Chapter 14: Problem 99

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.5.6 \times 10^{-4}\right)\)

Short Answer

Expert verified
The pH of the 0.20-M ethylamine solution is approximately 12.02. To obtain this value, we calculated the concentration of OH- ions using the given Kb value and the ICE table method, found the pOH, and then used the relationship between pH and pOH.

Step by step solution

01

Write the ionization reaction for ethylamine.

Ethylamine (C2H5NH2) is a weak base and it will react with water to form its conjugate acid (C2H5NH3+) and hydroxide ion (OH-). The reaction can be written as: C2H5NH2 + H2O ⟶ C2H5NH3+ + OH-
02

Write the Kb expression.

Using the ionization reaction, we can write the Kb expression for ethylamine as: Kb = \(\frac{[C_{2}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{2}H_{5}NH_{2}]}\)
03

Set up an ICE table.

To find the concentrations of the reacting species, we can create an ICE (Initial, Change, Equilibrium) table as follows: | | C2H5NH2 | C2H5NH3+ | OH- | |--------------|---------|----------|------| | Initial | 0.20 M | 0 M | 0 M | | Change | -x | +x | +x | | Equilibrium | 0.20-x | x | x |
04

Substitute the equilibrium concentrations in the Kb expression.

Substituting the equilibrium concentrations from the ICE table into the Kb expression: \(5.6 \times 10^{-4}\) = \(\frac{x \cdot x}{0.20-x}\)
05

Make an assumption to simplify the equation.

Since ethylamine is a weak base, we can assume that the change in its concentration (x) is small compared to its initial concentration (0.20 M). So, we have: 0.20 - x ≈ 0.20 Now, our equation becomes: \(5.6 \times 10^{-4}\) = \(\frac{x \cdot x}{0.20}\)
06

Solve for x to find the [OH-] concentration.

Solving the equation for x: x^2 = \(5.6 \times 10^{-4} \cdot 0.20\) x^2 = \(1.12 \times 10^{-4}\) x ≈ \(\sqrt{1.12 \times 10^{-4}}\) x ≈ \(1.06 \times 10^{-2}\) So, [OH-] = \(1.06 \times 10^{-2} M\)
07

Calculate pOH.

Now, we can calculate the pOH using the formula: pOH = -log([OH-]) pOH = -log(\(1.06 \times 10^{-2}\)) pOH ≈ 1.98
08

Calculate pH of the solution.

Finally, we can use the relationship between pH and pOH to find the pH of the solution: pH = 14 - pOH pH = 14 - 1.98 pH ≈ 12.02 The pH of the 0.20-M ethylamine solution is approximately 12.02.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of the following solutions. a. \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(1.0 \times 10^{-10} M \mathrm{NaOH}\) c. \(2.0 \mathrm{M} \mathrm{NaOH}\)

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