Question bank — Acid-base reactions — neutralization, salt formation

False. pH 7 only holds for a strong acid + strong base. If either parent is weak, the salt's ion reacts with water (Salt hydrolysis) and shifts the pH away from 7.

True here, because both are monoprotic/monobasic ($n_a=n_b=1$), so equal moles of $H^{+}$ and $O{H}^{-}$ meet. It is not generally true — see the diprotic case below.

False. $H_{2}S{O}_4$ supplies 2 $H^{+}$ per formula unit, so it needs twice the moles of $O{H}^{-}$; you'd need double the NaOH volume.

False. Forming the very stable O–H bonds of water is the whole energetic driver ($\Delta H\approx -57.3$ kJ/mol). The spectators are the salt's ions, which don't chemically change.

False. Only for strong acid + strong base, where the sole reaction is $H^{+}+O{H}^{-}\to H_{2}O$. With a weak acid/base, extra energy is spent breaking it apart first, so $\Delta H$ is less negative.

False. "Neutral" refers to overall charge balance, not pH. Many salts (e.g. $C{H}_{3}COONa$, $N{H}_{4}Cl$) give distinctly non-7 solutions.

False. It's the salt of a strong acid (HCl) and weak base ($N{H}_3$). The $N{H}_4^{+}$ cation hydrolyzes, releasing $H^{+}$, so the solution is acidic.

True. That's the point — for strong acid + strong base only the proton and hydroxide react; $N{a}^{+}$, $C{l}^{-}$, $K^{+}$ etc. are spectators and cancel out.

The equation is unbalanced. Sulfuric acid has 2 $H^{+}$ needing 2 $O{H}^{-}$, so it's $H_{2}S{O}_4+2KOH\to K_{2}S{O}_4+2H_{2}O$.

Wrong anion. The anion comes from the acid intact: nitric acid $HN{O}_3$ gives nitrate $N{O}_3^{-}$, so the salt is $NaN{O}_3$ (sodium nitrate), not nitrite.

Moles depend on molarity and the $n$ factor, not volume alone. The correct condition is $n_a{M}_a{V}_a=n_b{M}_b{V}_b$ (Molarity and solution stoichiometry).

Backwards. Acetate is the conjugate base of weak acetic acid; it grabs $H^{+}$ from water and releases $O{H}^{-}$, making the solution basic.

No neutralization happens between those two — they're spectators. The neutralization is strictly $H^{+}+O{H}^{-}\to H_{2}O$; $N{a}^{+}$ and $C{l}^{-}$ just stay dissolved.

The equivalence point (moles $H^{+}$ = moles $O{H}^{-}$) is fixed by chemistry. The indicator only signals an endpoint near it (Titration and indicators); a good indicator's endpoint ≈ equivalence point.

The acid's H is bonded to an electronegative atom that hogs electrons, leaving H electron-poor ($\delta +$); the base offers electron-rich lone pairs / $O{H}^{-}$. Electron-poor is pulled toward electron-rich, so the proton transfers.

Because after dissociation the only chemical change is always the same reaction, $H^{+}+O{H}^{-}\to H_{2}O$. The spectator ions were already free before and after, contributing nothing to $\Delta H$ (Enthalpy of reaction).

Strong electrolytes dissociate essentially 100% in water, so they exist as free ions, not molecules. Writing them as ions lets the true spectators cancel, exposing the real reaction.

The anion is the conjugate base of a weak acid, so it's a decent proton-grabber. It pulls $H^{+}$ off water, leaving excess $O{H}^{-}$ behind → basic solution.

Forming the strong, stable O–H bonds of water releases more energy than was needed to bring $H^{+}$ and $O{H}^{-}$ together — a net drop to lower energy, released as heat (exothermic).

Sulfuric acid is diprotic — one formula unit can donate two $H^{+}$. Each $H^{+}$ needs its own $O{H}^{-}$ partner to become water, so two hydroxides are required.

Above 7 (basic). All the acid has become its conjugate-base salt, which hydrolyzes to give $O{H}^{-}$ — so the equivalence point is not at pH 7.

The leftover, un-reacted $O{H}^{-}$. Past the equivalence point there's no more $H^{+}$ to consume, so pH is set by the excess strong base and climbs high.

Two equivalence points, one after each proton is removed, because the two ionizable $H^{+}$ come off in separate steps at different concentrations.

They still neutralize, but the resulting salt has both ions hydrolyzing. The final pH depends on which parent was relatively stronger; it may be near, above, or below 7.

Neither in the acid-base sense — $NaCl$ is the salt of a strong acid and strong base, so neither ion hydrolyzes and the solution stays neutral (pH ≈ 7). No net $H^{+}/O{H}^{-}$ reaction occurs.

Water is neutral because $[H^{+}]=[O{H}^{-}]$, but the amounts are minuscule ($10^{-7}$ M each). "Neutralization" refers to deliberately balancing added acid and base, not this trace self-ionization.