Exercise 1.8 1. For a 0.040 M ammonia solution: a. Write a chemical equation that shows the ionization of ammonia in water. b. Calculate the equilibrium concentration of ammonia, ammonium ions and hydroxide ions, using the approximation method. Check whether the approximation is valid or not. c. If the approximation is invalid, use the quadratic formula to calculate the concentration of ammonia, ammonium ions and hydroxide ions. d. Calculate the pOH and pH of the solution.

### Exercise 1.8 #### 1. For a 0.040 M ammonia solution: **a. Write a chemical equation that shows the ionization of ammonia in water.** Ammonia (\( \text{NH}_3 \)) ionizes in water as follows: \[ \text{NH}_3 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{NH}_4^+ (aq) + \text{OH}^- (aq) \] **b. Calculate the equilibrium concentration of ammonia, ammonium ions, and hydroxide ions, using the approximation method. Check whether the approximation is valid or not.** Let \( x \) be the change in concentration of \( \text{NH}_4^+ \) and \( \text{OH}^- \) at equilibrium. Initial concentrations: - \( [\text{NH}_3] = 0.040 \, \text{M} \) - \( [\text{NH}_4^+] = 0 \) - \( [\text{OH}^-] = 0 \) At equilibrium: - \( [\text{NH}_3] = 0.040 - x \) - \( [\text{NH}_4^+] = x \) - \( [\text{OH}^-] = x \) The equilibrium constant \( K_b \) for ammonia is approximately \( 1.8 \times 10^{-5} \). Using the expression for \( K_b \): \[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = \frac{x \cdot x}{0.040 - x} = \frac{x^2}{0.040 - x} \] Assuming \( x \) is small compared to 0.040 M, we can approximate \( 0.040 - x \approx 0.040 \): \[ 1.8 \times 10^{-5} = \frac{x^2}{0.040} \] Now, solving for \( x \): \[ x^2 = 1.8 \times 10^{-5} \times 0.040 \] \[ x^2 = 7.2 \times 10^{-7} \] \[ x = \sqrt{7.2 \times 10^{-7}} \approx 8.49 \times 10^{-4} \, \text{M} \] Now we can find the equilibrium concentrations: - \( [\text{NH}_3] = 0.040 - x \approx 0.040 - 0.000849 \approx 0.03915 \, \text{M} \) - \( [\text{NH}_4^+] = x \approx 8.49 \times 10^{-4} \, \text{M} \) - \( [\text{OH}^-] = x \approx 8.49 \times 10^{-4} \, \text{M} \) **Check the validity of the approximation:** To check if the approximation is valid, we need to see if \( x \) is less than 5% of the initial concentration of ammonia: \[ \frac{x}{0.040} \times 100\% = \frac{8.49 \times 10^{-4}}{0.040} \times 100\% \approx 2.12\% \] Since 2.12% is less than 5%, the approximation is valid. **c. If the approximation is invalid, use the quadratic formula to calculate the concentration of ammonia, ammonium ions, and hydroxide ions.** Since the approximation is valid, we do not need to use the quadratic formula in this case. **d. Calculate the pOH and pH of the solution.** To find the pOH: \[ \text{pOH} = -\log[\text{OH}^-] = -\log(8.49 \times 10^{-4}) \approx 3.07 \] To find the pH: Using the relationship \( \text{pH} + \text{pOH} = 14 \): \[ \text{pH} = 14 - \text{pOH} = 14 - 3.07 \approx 10.93 \] ### Summary of Results: - \( [\text{NH}_3] \approx 0.03915 \, \text{M} \) - \( [\text{NH}_4^+] \approx 8.49 \times 10^{-4} \, \text{M} \) - \( [\text{OH}^-] \approx 8.49 \times 10^{-4} \, \text{M} \) - \( \text{pOH} \approx 3.07 \) - \( \text{pH} \approx 10.93 \)