Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 M \mathrm{KNO}_{2}\) b. \(0.45 M\) NaOCl c. \(0.40 M \mathrm{NH}_{4} \mathrm{ClO}_{4}\)

We first need to identify the salt and the ions present in the given solution. The salt in the solution is potassium nitrite (\(\mathrm{KNO}_{2}\)), which dissociates into \(\mathrm{K}^+\) and \(\mathrm{NO}_{2}^{-}\) ions. Since \(\mathrm{K}^+\) ions are from a strong base (\(\mathrm{KOH}\)) and \(\mathrm{NO}_{2}^{-}\) ions are from a weak acid (\(\mathrm{HNO}_{2}\)), the salt can be considered as the conjugate base in this solution. To determine the pH, we start by writing the ionization reaction of \(\mathrm{NO}_{2}^{-}\): \[\mathrm{NO}_{2}^{-} + H_{2}O \rightleftharpoons \mathrm{HNO}_{2} + OH^{-}\] Next, we need to determine the base ionization constant (\(K_{b}\)) for this reaction. Knowing the acid ionization constant of \(\mathrm{HNO}_{2}\) (\(K_{a} = 4.5 \times 10^{-4}\)), we can use the relationship \(K_{a}\times K_{b} = K_{w}\), where \(K_{w}\) is the ion product for water \((1.0 \times 10^{-14})\). \[ K_{b} = \frac{K_{w}}{K_{a}}\] Now, we can use the given initial concentration of \(\mathrm{KNO}_{2}\) and solve for the concentration of \(OH^-\) ions using an equilibrium table. Once we have the concentration of \(OH^-\) ions, we can calculate the \(pOH\) using \(pOH = -\log_{10}[OH^-]\), and finally, we can find the pH using the relationship \(pH = 14 - pOH\).

The salt in the given solution is sodium hypochlorite (NaOCl), which dissociates into Na\(^+\) and OCl\(^-\) ions. Since Na\(^+\) ions are from a strong base (NaOH) and OCl\(^-\) ions are from a weak acid (HClO), NaOCl can be considered as a weak base. To determine the pH, we start by writing the ionization reaction of OCl\(^-\): \[ \mathrm{OCl}^{-} + H_{2}O \rightleftharpoons \mathrm{HClO} + OH^{-}\] We need to determine the base ionization constant (\(K_{b}\)) for this reaction. Knowing the acid ionization constant of \(\mathrm{HClO}\) (\(K_{a} = 3.0 \times 10^{-8}\)), we can use the relationship \(K_{a}\times K_{b} = K_{w}\). \[ K_{b} = \frac{K_{w}}{K_{a}}\] Now, we will solve for the concentration of \(OH^-\) ions using the equilibrium table with the initial concentration of NaOCl given. Once we have the concentration of \(OH^-\) ions, we can calculate the \(pOH\) using \(pOH = -\log_{10}[OH^-]\), and finally, we can find the pH using the relationship \(pH = 14 - pOH\).

The salt in the given solution is ammonium perchlorate (\(\mathrm{NH}_{4} \mathrm{ClO}_{4}\)), which dissociates into \(\mathrm{NH}_{4}^+\) and \(\mathrm{ClO}_{4}^-\) ions. Since \(\mathrm{NH}_{4}^+\) ions are from a weak base (\(\mathrm{NH}_{3}\)) and \(\mathrm{ClO}_{4}^{-}\) ions are from a strong acid (\(\mathrm{HClO}_{4}\)), the salt can be considered as the conjugate acid in this solution. To determine the pH, we start by writing the ionization reaction of \(\mathrm{NH}_{4}^+\): \[ \mathrm{NH}_{4}^{+} + H_{2}O \rightleftharpoons \mathrm{NH}_{3} + H_{3}O^{+}\] We need to determine the acid ionization constant (\(K_{a}\)) for this reaction. Knowing the base ionization constant of \(\mathrm{NH}_{3}\) (\(K_{b} = 1.8 \times 10^{-5}\)), we can use the relationship \(K_{a} \times K_{b} = K_{w}\). \[ K_{a} = \frac{K_{w}}{K_{b}}\] Now, we will solve for the concentration of \(H_{3}O^+\) ions using the equilibrium table with the initial concentration of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) given. Once we have the concentration of \(H_{3}O^+\) ions, we can calculate the pH using \(pH = -\log_{10}[H_{3}O^{+}]\).