Is an aqueous solution of NaHSO_ acidic, basic, or neutral? What reaction occurs with water? Calculate the pH of a \(0.10-M\) solution of NaHSO.

When NaHSO is dissolved in water, it dissociates into Na and HSO ions, represented by the following equation: \( NaHSO_{3} \rightarrow Na^{+} + HSO_{3}^{-} \)

To determine if the solution is acidic, basic, or neutral, we need to check if the HSO ion acts as an acid or a base. HSO can act as an acid by donating a proton (H) to a water molecule, which forms a hydronium ion (H3O+) and SO ions, according to the following equilibrium reaction: \( HSO_{3}^{-} + H_{2}O \leftrightarrow H_{3}O^{+} + SO_{3}^{2-} \) HSO can also act as a base by accepting a proton (H) from a water molecule, which forms a hydroxide ion (OH-) and HSO ions, according to the equilibrium reaction: \( HSO_{3}^{-} + H_{2}O \leftrightarrow OH^{-} + H_{2}SO_{3} \) Since HSO can both donate and accept protons, it is behaving as an amphoteric ion. Consequently, the solution has both acidic and basic properties.

To determine if the solution is overall acidic or basic, we need to compare the acid dissociation constant (Ka) of HSO with its base dissociation constant (Kb). For HSO, Ka = 1.7 x 10, and the Kb = Kw / Ka, where Kw = \(1.0 x 10^{-14}\). Calculating Kb for HSO: \(Kb = \frac{1.0 \times 10^{-14}}{1.7 \times 10^{-7}} = 5.9 \times 10^{-8}\) Since Ka > Kb, the acidic reaction is dominant, making the solution acidic.

To determine the pH of a 0.10 M NaHSO solution, we will use the Ka value and the initial concentration of NaHSO. The dissociation reaction of HSO is as follows: \( HSO_{3}^{-} \leftrightarrow H^{+} + SO_{3}^{2-} \) The initial concentration of HSO is 0.10 M, and the concentrations of Hand SO are both initially 0. Applying the equilibrium expression: \( K_a = 1.7 \times 10^{-7} = \frac{[H^{+}][SO_{3}^{2-}]}{[HSO_{3}^{-}]} \) Let x be the concentration of Hand SO at equilibrium. Then, the Ka expression becomes: \( 1.7 \times 10^{-7} = \frac{x^2}{0.10 - x} \) Assuming that x << 0.10, we can approximately say: \( 1.7 \times 10^{-7} = \frac{x^2}{0.10} \) Solving for x (the concentration of H ions) gives: \( x = \sqrt{1.7 \times 10^{-7} \times 0.10} = 1.3 \times 10^{-4} \) Finally, the pH is determined using the formula: \( pH = -\log_{10} [H^{+}] \) \( pH = -\log_{10} (1.3 \times 10^{-4}) = 3.89 \) Hence, the pH of a 0.10 M NaHSO solution is 3.89, confirming that it is an acidic solution.