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When sodium bisulfate (NaHSO4) dissolves in water, it forms the following ions: \[ \mathrm{NaHSO}_{4(s)} \rightarrow \mathrm{Na^+_{(aq)}} + \mathrm{HSO^{-}_{4(aq)}} \] The sodium ion (Na+) is a neutral ion, as it does not have the ability to act as either an acid or a base. However, the bisulfate ion (HSO4-) can act as either an acid (by donating a proton, H+) or a base (by accepting a proton, H+).

Based on the acidic dissociation constant of sulfuric acid (H2SO4), which is around \(K_{a1}=10^3\), it can lose a proton easily, resulting in HSO4- which has a lower Ka around \(K_{a2}=1.2 \times 10^{-2}\). Since HSO4-has a significant Ka value, it will act as a weak acid in water, as shown in the following reaction: \[ \mathrm{HSO^{-}_{4(aq)}} + \mathrm{H}_{2}\mathrm{O_{(l)}} \rightleftharpoons \mathrm{H}_{3}\mathrm{O^{+}_{(aq)}} + \mathrm{SO^{2-}_{4(aq)}} \] As HSO4- acts as an acid, the reaction with water will result in an acidic solution.

For the equilibrium constant\(K_{a2}\) for the reaction above, we can write: \[ K_{a2} = \frac{ [\mathrm{H}_{3}\mathrm{O^+}][\mathrm{SO^{2-}_{4} }] }{[\mathrm{HSO^{-}_{4}]} \]

At equilibrium, we assume the initial concentration of H3O+ ions and sulfate ions (SO42-) to be zero. Since the NaHSO4 solution has an initial concentration of 0.10 M, we can write the following changes in concentration at equilibrium: - \([\mathrm{HSO^{-}_{4}}] = 0.10 - x\) - \([\mathrm{H}_{3}\mathrm{O^+}] = x\) - \([\mathrm{SO^{2-}_{4}}] = x\) Plugging these values in the Ka expression, we get: \(1.2 \times 10^{-2} = \frac{x^2}{0.10 - x}\) Assuming that the value of x is much smaller than 0.10, we can simplify the equation to: \(1.2\times10^{-2} \approx \frac{x^2}{0.10}\) Now, solve for x: \(x^2 = 1.2\times10^{-2} \times 0.10\) \(x \approx \sqrt{1.2\times10^{-3}}\) \(x \approx 3.5\times10^{-2}\) Therefore, the concentration of H3O+ ions at equilibrium is \(\approx 3.5\times10^{-2} \,\mathrm{M}\).

Now that we have the concentration of H3O+ ions, we can calculate the pH using the formula: \[ \mathrm{pH} = -\log{[\mathrm{H_{3}O^{+}}]} \] \[ \mathrm{pH} = -\log{(3.5\times10^{-2})} \] \[ \mathrm{pH} \approx 1.5 \] The pH of a 0.10 M solution of NaHSO4 is approximately 1.5, which confirms that the aqueous solution of NaHSO4 is acidic.