U4.8 · Introduction to Acid-Base ReactionsPractice · 酸碱反应入门

A Brønsted–Lowry acid donates $\ce{H+}$. In the forward reaction $\ce{H2O}$ gives an $\ce{H+}$ to $\ce{CH3NH2}$ and becomes $\ce{OH-}$, so $\ce{H2O}$ is the acid and $\ce{CH3NH2}$ the base (like $\ce{NH3}$, its N lone pair accepts the $\ce{H+}$). $\ce{CH3NH3+}$ and $\ce{OH-}$ are products: the conjugate acid and conjugate base.

A conjugate base is what remains after an acid loses one $\ce{H+}$: one fewer H and a charge one lower. $\ce{HPO4^2-}$ minus $\ce{H+}$ is $\ce{PO4^3-}$. $\ce{H2PO4-}$ is its conjugate acid (one more $\ce{H+}$), $\ce{H3PO4}$ has two more, and $\ce{OH-}$ is unrelated.

n(KOH) = 0.400 × 0.0300 = 0.0120 mol. $\ce{H2SO4}$ is diprotic and reacts with KOH 1 : 2, so n($\ce{H2SO4}$) = 0.00600 mol and V = 0.00600 ÷ 0.250 = 0.0240 L = 24.0 mL. A 1 : 1 ratio would give 48.0 mL.

A conjugate pair differs by exactly one $\ce{H+}$. $\ce{H2SO4}$ and $\ce{SO4^2-}$ differ by two, with $\ce{HSO4-}$ in between: the conjugate base of $\ce{H2SO4}$ is $\ce{HSO4-}$, and so is the conjugate acid of $\ce{SO4^2-}$. The other three pairs each differ by one $\ce{H+}$.

A base accepts $\ce{H+}$. In B, $\ce{H2O}$ takes an $\ce{H+}$ from $\ce{HNO2}$ and becomes $\ce{H3O+}$: it acts as a base. In A, C, and D, $\ce{H2O}$ gives an $\ce{H+}$ to $\ce{NH3}$, $\ce{CN-}$, or $\ce{HCO3-}$ and becomes $\ce{OH-}$: it acts as an acid. Whether water ends up as $\ce{H3O+}$ or $\ce{OH-}$ tells you which.