15.37 For a 0.45 M solution of $$ \mathrm{Na}_{2} \mathrm{SO}_{3} $$, do the following: (a) Identify the major species. (b) Identify the equilibrium that determines the pH. (c) Compute the pH .

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**(a) Identify the major species** When $$\ce{Na2SO3}$$ dissolves, it dissociates completely into sodium and sulfite ions: $$ \ce{Na2SO3 -> 2Na^+ + SO3^{2-}} $$ Thus, the major species in a 0.45 M $$\ce{Na2SO3}$$ solution are: - $$\ce{Na^+}$$ (concentration: $$2 \times 0.45 = 0.90$$ M) - $$\ce{SO3^{2-}}$$ (concentration: 0.45 M) **(b) Identify the equilibrium that determines the pH** The $$\ce{SO3^{2-}}$$ ion is the conjugate base of $$\ce{HSO3^-}$$ (which comes from the diprotic acid $$\ce{H2SO3}$$). In water, $$\ce{SO3^{2-}}$$ undergoes hydrolysis: $$ \ce{SO3^{2-} + H2O <=> HSO3^- + OH^-} $$ This equilibrium is responsible for generating $$\ce{OH^-}$$ and, hence, making the solution basic. It is the equilibrium that determines the pH. **(c) Compute the pH** 1. **Determine the base ionization constant, $$K_b$$:** The hydrolysis reaction for $$\ce{SO3^{2-}}$$ has an equilibrium constant $$K_b$$ related to the second acid dissociation constant $$K_{a2}$$ of $$\ce{H2SO3}$$ by: $$ K_b = \frac{K_w}{K_{a2}} $$ where $$K_w = 1.0 \times 10^{-14}$$ and typically $$pK_{a2} \approx 7.2$$, giving: $$ K_{a2} = 10^{-7.2} \approx 6.31 \times 10^{-8}. $$ Then, $$ K_b = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-8}} \approx 1.585 \times 10^{-7}. $$ 2. **Set up the hydrolysis equilibrium:** Let $$x$$ be the amount of $$\ce{SO3^{2-}}$$ that reacts: $$ \begin{array}{rcl} \ce{SO3^{2-}} & \ce{+ H2O} & \ce{<=> HSO3^- + OH^-} \ \text{Initial:} & 0.45 & 0 \quad\quad 0 \ \text{Change:} & -x & +x \quad +x \ \text{Equilibrium:} & 0.45 - x & x \quad\quad x \ \end{array} $$ The equilibrium expression is: $$ K_b = \frac{[ \ce{HSO3^-} ][ \ce{OH^-} ]}{[ \ce{SO3^{2-}} ]} = \frac{x^2}{0.45 - x}. $$ 3. **Assume $$x \ll 0.45$$ and solve for $$x$$:** Under the assumption, $$0.45 - x \approx 0.45$$, so: $$ x^2 \approx K_b \times 0.45. $$ Substituting the value of $$K_b$$: $$ x^2 \approx 1.585 \times 10^{-7} \times 0.45 \approx 7.1325 \times 10^{-8}. $$ Taking the square root: $$ x \approx \sqrt{7.1325 \times 10^{-8}} \approx 8.45 \times 10^{-4}. $$ This $$x$$ represents the equilibrium concentration of $$\ce{OH^-}$$. 4. **Calculate the pOH and pH:** $$ \text{pOH} = -\log [\ce{OH^-}] = -\log (8.45 \times 10^{-4}) \approx 3.07. $$ Finally, the pH is: $$ \text{pH} = 14 - \text{pOH} \approx 14 - 3.07 \approx 10.93. $$ **Answer Summary:** - Major species: $$\ce{Na^+}$$ (0.90 M) and $$\ce{SO3^{2-}}$$ (0.45 M). - pH-determining equilibrium: $$\ce{SO3^{2-} + H2O <=> HSO3^- + OH^-}$$. - The pH of the solution is approximately $$10.93$$. **(a) Major Species:** - Sodium ions ($$\ce{Na^+}$$) at 0.90 M - Sulfite ions ($$\ce{SO3^{2-}}$$) at 0.45 M **(b) pH-Determining Equilibrium:** $$ \ce{SO3^{2-} + H2O <=> HSO3^- + OH^-} $$ **(c) pH Calculation:** The pH of the solution is approximately **10.93**.

15.37 For a 0.45 M solution of \( \mathrm{Na}_{2} \mathrm{SO}_{3} \), do the following: (a) Identify the major species. (b) Identify the equilibrium that determines the pH. (c) Compute the pH .

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