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Chapter 16: Problem 60

Determine the \(\mathrm{pH}\) of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix D): (a) \(0.095 \mathrm{M}\) hypochlorous acid, \((\mathbf{b}) 0.0085 \mathrm{M}\) hydrazine, (c) \(0.165 \mathrm{M}\) hydroxylamine.

Short Answer

Expert verified
Using the given initial concentrations and equilibrium constant expressions, the pH of each solution can be determined as follows: (a) Hypochlorous acid (acidic): Calculate [H+] using \(K_a = \frac{x^2}{0.095-x}\), then find pH: \(pH = -\log([H^+])\) (b) Hydrazine (basic): Calculate [OH-] using \(K_b = \frac{x^2}{0.0085-x}\), find pOH: \(pOH = -\log([OH^-])\), and convert to pH: \(pH = 14 - pOH\) (c) Hydroxylamine (basic): Calculate [OH-] using \(K_b = \frac{x^2}{0.165-x}\), find pOH: \(pOH = -\log([OH^-])\), and convert to pH: \(pH = 14 - pOH\)

Step by step solution

01

Identify whether the solution is acidic or basic

In this exercise, three solutions are given: (a) Hypochlorous acid: As the name suggests, it is an acid. (b) Hydrazine: This is a base. (c) Hydroxylamine: This is a base.
02

Write down the ionization/dissociation reactions for each solution

For each of the given solutions, we need to write down their ionization or dissociation reactions, as follows: (a) Hypochlorous acid: \(HOCl \rightleftharpoons H^+ + OCl^-\) (b) Hydrazine: \(N_2H_4 + H_2O \rightleftharpoons N_2H_5^+ + OH^-\) (c) Hydroxylamine: \(NH_2OH + H_2O \rightleftharpoons NH_3OH^+ + OH^-\)
03

Set up an equilibrium table and expressions for each solution

Now, we will set up an equilibrium table for each of these solutions, as well as the equilibrium constant expressions: (a) Hypochlorous acid: \(K_a = \frac{[H^+][OCl^-]}{[HOCl]}\) Initial concentration: [HOCl] = 0.095 M, [H+] = 0, [OCl-] = 0 (b) Hydrazine: \(K_b = \frac{[N_2H_5^+][OH^-]}{[N_2H_4]}\) Initial concentration: [N_2H_4] = 0.0085 M, [N_2H_5+] = 0, [OH-] = 0 (c) Hydroxylamine: \(K_b = \frac{[NH_3OH^+][OH^-]}{[NH_2OH]}\) Initial concentration: [NH_2OH] = 0.165 M, [NH_3OH+] = 0, [OH-] = 0
04

Calculate [H+] or [OH-] using equilibrium constants

For each solution, refer to Appendix D and use the given equilibrium constant to find the concentration of [H+] or [OH-]. Solve the equations assuming x as the change in concentration: (a) Hypochlorous acid: As there is no initial [H+] or [OCl-], let x be the change. At equilibrium, [H+] = x, [OCl-] = x, and [HOCl] = 0.095 - x: \(K_a = \frac{x^2}{0.095-x}\) (b) Hydrazine: At equilibrium, [N_2H_5+] = x, [OH-] = x, and [N_2H_4] = 0.0085 - x: \(K_b = \frac{x^2}{0.0085-x}\) (c) Hydroxylamine: At equilibrium, [NH_3OH+] = x, [OH-] = x, and [NH_2OH] = 0.165 - x: \(K_b = \frac{x^2}{0.165-x}\)
05

Determine the pH/pOH of the solutions

Calculate the [H+] or [OH-] concentration for each solution by solving the respective equilibrium constant expression. Then, find the pH or pOH as appropriate: (a) Hypochlorous acid: Calculate [H+], then find the pH: \(pH = -\log([H^+])\) (b) Hydrazine: Calculate [OH-], then find the pOH: \(pOH = -\log([OH^-])\) (c) Hydroxylamine: Calculate [OH-], then find the pOH: \(pOH = -\log([OH^-])\)
06

Convert pOH to pH if needed

If you have calculated the pOH in Step 5, convert it to pH using the relationship: \(pH + pOH = 14\) (a) Hypochlorous acid: No conversion needed; the pH is already calculated. (b) Hydrazine: Convert the pOH to pH. (c) Hydroxylamine: Convert the pOH to pH. Following these steps, you will find the pH of each given solution.

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Most popular questions from this chapter

Calculate the pH of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix D): (a) \(0.150 \mathrm{M}\) propionic acid $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)$ (b) \(0.250 \mathrm{M}\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),(\mathbf{c}) 0.750 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\)

Determine whether each of the following is true or false: (a) All strong bases are salts of the hydroxide ion. (b) The addition of a strong base to water produces a solution of \(\mathrm{pH}>7.0\) (c) Because \(\mathrm{Mg}(\mathrm{OH})_{2}\) is not very soluble, it cannot be a strong base.

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and $\mathrm{H}_{3} \mathrm{NOH}^{+}$.

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine, $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;$ (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}$

Calculate the pH of each of the following strong acid solutions: $(\mathbf{a}) 0.0178 \mathrm{M} \mathrm{HNO}_{3},(\mathbf{b}) 0.500 \mathrm{~g}$ of \(\mathrm{HClO}_{3}\) in \(5.00 \mathrm{~L}\) of solution, \((\mathbf{c}) 5.00 \mathrm{~mL}\) of $2.00 \mathrm{M} \mathrm{HCl}\( diluted to \)0.500 \mathrm{~L}$, (d) a mixture formed by adding \(75.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HBr}\).
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