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

For each of the solutions, we will first determine the concentration of their respective conjugate acid/base, required for the calculation of pH. For Solution a: \(\mathrm{KNO}_{2}\) dissociates in water to form \(\mathrm{OH}^{-}\) ions and \(\mathrm{NO}_{2}^{-}\) ions. Since the concentration of \(\mathrm{KNO}_{2}\) is given, the concentration of \(\mathrm{NO}_{2}^{-}\) ions will also be \(0.12\)M. For Solution c: \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) dissociates in water to form \(\mathrm{NH}_{4}^{+}\) ions and \(\mathrm{ClO}_{4}^{-}\) ions. The concentration of \(\mathrm{NH}_{4}^{+}\) ions will be \(0.40\)M. For Solution b: \(\mathrm{NaOCl}\) dissociates in water to form \(\mathrm{Na}^{+}\) ions and \(\mathrm{OCl}^{-}\) ions. The concentration of \(\mathrm{OCl}^{-}\) ions will be \(0.45\)M.

We will now find the concentration of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions in each solution using the equilibrium constants (Ka or Kb) of their respective conjugate acid/base. For Solution a: The equilibrium constant for the dissociation of \(\mathrm{NO}_{2}^{-}\) ions, Kb, is given by: \[K_b = \frac{[\mathrm{HNO}_{2+}][\mathrm{OH}^−]}{[\mathrm{NO}_{2}^{−}]}\] The \(\mathrm{OH}^{-}\) concentration can be calculated as: \[\mathrm{OH}^{-} = \sqrt{K_b \times [\mathrm{NO}_{2}^{-}]}\] For Solution c: The equilibrium constant for the dissociation of \(\mathrm{NH}_{4}^{+}\) ions, Ka, is given by: \[K_a = \frac{[\mathrm{NH}_{3}][\mathrm{H}^+]}{[\mathrm{NH}_{4}^{+}]}\] The \(\mathrm{H}^{+}\) concentration can be calculated as: \[\mathrm{H}^{+} = \sqrt{K_a \times [\mathrm{NH}_{4}^{+}]}\] For Solution b: The equilibrium constant for the dissociation of \(\mathrm{OCl}^{-}\) ions, Kb, is given by: \[K_b = \frac{[\mathrm{HOCl+}][\mathrm{OH}^-]}{[\mathrm{OCl}^{-}]}\] The \(\mathrm{OH}^{-}\) concentration can be calculated as: \[\mathrm{OH}^{-} = \sqrt{K_b \times [\mathrm{OCl}^{-}]}\]

Finally, we will calculate the pH for each solution using the \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) concentrations. For Solution a: Calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \[pOH = -\log([\mathrm{OH}^{-}])\] Then, calculate pH using the relationship: \[pH = 14 - pOH\] For Solution c: Calculate pH using the \(\mathrm{H}^{+}\) concentration: \[pH = -\log([\mathrm{H}^+])\] For Solution b: Calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \[pOH = -\log([\mathrm{OH}^{-}])\] Then, calculate pH using the relationship: \[pH = 14 - pOH\]