Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) HONH \(_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 M\) HONH \(_{3}\) Cl c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing 0.100 \(M \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

This is a weak base, so we will use the Kb (base dissociation constant) to establish a reaction for the base in water: \( HONH_{2} + H_{2}O \rightleftharpoons OH^- + HONH_{3} \) Now, let's create the ICE (Initial, Change, Equilibrium) table for this reaction: | Base | -OH | +HONH3 | |---------|-----|--------| | Initial |0.100 M| 0 |0 | | Change |-x | +x | +x | | Equilibrium |0.100-x | x | x | \[ Kb = \dfrac{[OH^-][HONH_{3}]}{[HONH_{2}]} = \dfrac{x^2}{0.100 - x} = 1.1 \times 10^{-8} \] Now we'll need to solve this equation for x. Since Kb is very small, we don't have to consider the effect of the decrease in the initial concentration of HONH2 and can assume x is very small compared to 0.100 M. Therefore, x²/0.100 = 1.1 × 10⁻⁸. To find the pH, let's determine the pOH first: \( pOH = -\log_{10} [OH^-] = -\log_{10}x \) Now, find the pH using the following formula: \( pH = 14 - pOH \)

This is a salt resulting from a weak base (HONH2) and a strong acid (HCl). When dissolved in water, it undergoes hydrolysis: \( HONH_{3}^{+} + H_{2}O \rightleftharpoons HONH_{2} + H_{3}O^{+} \) Set up either Kb or Ka since HONH3 is the conjugate acid of HONH2 (our weak base in part a). Using the relationship: \( K_{a} \times K_{b} = K_{w} \) where \( K_{w} = 1 \times 10^{-14} \) We have Kb, so let's find Ka: \( K_{a} = \dfrac{K_w}{K_b} = \dfrac{1 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.1 \times 10^{-7} \) Now, set up the ICE table: | Acid | -H3O | +HONH2 | |----------|-----|--------| | Initial |0.100 M| 0 |0 | | Change |-x | +x | +x | | Equilibrium | 0.100-x| x | x| \[ K_{a} = \dfrac{[H_{3}O^{+}][HONH_{2}]}{[HONH_{3}^+]} = \dfrac{x^2}{0.100 - x} = 9.09 \times 10^{-7} \] Solving for x: \( x^2 = (0.100) \times (9.1 \times 10^{-7}) \) Now, we'll find the pH using: \( pH = -\log_{10} [H_3O^+] = -\log_{10}(x) \)

This is pure water, so the concentrations of H3O+ and OH- are equal: \[ [H_{3}O^{+}] = [OH^{-}] \] Since the product of their concentrations equals the ion product constant for water, Kw: \[ K_{w} = [H_{3}O^{+}][OH^{-}] = 1 \times 10^{-14} \] Because [H3O+] = [OH-]: \[ [H_{3}O^{+}]^2 = 1 \times 10^{-14} \] Taking the square root: \[ [H_{3}O^{+}] = 1 \times 10^{-7} \] And finally, find the pH: \[ pH = -\log_{10}[H_{3}O^{+}] = 7 \]

This is a buffer solution, as it contains a weak base and its conjugate acid. The best way to approach this problem is by using the Henderson-Hasselbalch equation: \[ pH = pK_{a} + \log \dfrac{[Base]}{[Acid]} \] We already have the value of Ka from part b. So, let's calculate pKa: \[ pK_{a} = -\log_{10}K_{a} \] Now, we plug in the concentrations of the base (HONH2) and the acid (HONH3+), and calculate the pH: \[ pH = pK_{a} + \log \dfrac{[HONH_{2}]}{[HONH_{3}^+]} \]