301. The relation defining \(pH\) is
ⓐ. \(pH=[\mathrm{H^+}]+[\mathrm{OH^-}]\)
ⓑ. \(pH=\frac{[\mathrm{OH^-}]}{[\mathrm{H^+}]}\)
ⓒ. \(pH=-\log[\mathrm{H^+}]\)
ⓓ. \(pH=\log[\mathrm{OH^-}]\)
Correct Answer: \(pH=-\log[\mathrm{H^+}]\)
Explanation: The \(pH\) scale is a logarithmic way to express hydrogen ion concentration. It is defined by \(pH=-\log[\mathrm{H^+}]\). Because of the negative sign, a higher \([\mathrm{H^+}]\) gives a lower \(pH\). This inverse relationship explains why acidic solutions have lower \(pH\) values. The concentration inside the logarithm is normally treated relative to the standard concentration convention used at this level.
302. A solution has \([\mathrm{H^+}]=1.0\times10^{-3}\,\text{M}\). Its \(pH\) is
ⓐ. \(7\)
ⓑ. \(3\)
ⓒ. \(11\)
ⓓ. \(14\)
Correct Answer: \(3\)
Explanation: \( \textbf{Definition:} \)
\[pH=-\log[\mathrm{H^+}]\]
\( \textbf{Given concentration:} \)
\[[\mathrm{H^+}]=1.0\times10^{-3}\,\text{M}\]
\( \textbf{Substitution:} \)
\[pH=-\log(1.0\times10^{-3})\]
\( \textbf{Log value:} \)
\[\log(1.0\times10^{-3})=-3\]
\( \textbf{Calculation:} \)
\[pH=-(-3)=3\]
\( \textbf{Final answer:} \) \(3\). A hydrogen ion concentration of \(10^{-3}\,\text{M}\) is acidic at \(298\,\text{K}\).
303. At \(298\,\text{K}\), the relation between \(pH\) and \(pOH\) is
ⓐ. \(pH-pOH=14\)
ⓑ. \(pH\times pOH=14\)
ⓒ. \(pH=pOH+14\)
ⓓ. \(pH+pOH=14\)
Correct Answer: \(pH+pOH=14\)
Explanation: At \(298\,\text{K}\), \(K_w=1.0\times10^{-14}\), so \(pK_w=14\). Since \(pH=-\log[\mathrm{H^+}]\) and \(pOH=-\log[\mathrm{OH^-}]\), the product relation \(K_w=[\mathrm{H^+}][\mathrm{OH^-}]\) becomes a sum relation. Therefore \(pH+pOH=pK_w=14\). The value \(14\) is tied to \(298\,\text{K}\), not to every temperature.
304. At \(298\,\text{K}\), a solution has \(pOH=5.0\). Its \(pH\) is
ⓐ. \(9.0\)
ⓑ. \(14.0\)
ⓒ. \(5.0\)
ⓓ. \(7.0\)
Correct Answer: \(9.0\)
Explanation: \( \textbf{Relation at \(298\,\text{K}\):} \)
\[pH+pOH=14\]
\( \textbf{Given value:} \)
\[pOH=5.0\]
\( \textbf{Solve for \(pH\):} \)
\[pH=14-pOH\]
\( \textbf{Substitution:} \)
\[pH=14-5.0\]
\( \textbf{Calculation:} \)
\[pH=9.0\]
\( \textbf{Final answer:} \) \(9.0\). Since the \(pH\) is greater than \(7\) at \(298\,\text{K}\), the solution is basic.
305. A solution at \(298\,\text{K}\) has \(pH=2\). The hydrogen ion concentration is
ⓐ. \(2.0\,\text{M}\)
ⓑ. \(1.0\times10^{-2}\,\text{M}\)
ⓒ. \(1.0\times10^{-7}\,\text{M}\)
ⓓ. \(1.0\times10^{-12}\,\text{M}\)
Correct Answer: \(1.0\times10^{-2}\,\text{M}\)
Explanation: \( \textbf{Definition:} \)
\[pH=-\log[\mathrm{H^+}]\]
\( \textbf{Given value:} \)
\[pH=2\]
\( \textbf{Convert from \(pH\) to concentration:} \)
\[[\mathrm{H^+}]=10^{-pH}\]
\( \textbf{Substitution:} \)
\[[\mathrm{H^+}]=10^{-2}\,\text{M}\]
\( \textbf{Final answer:} \) \(1.0\times10^{-2}\,\text{M}\). A lower \(pH\) corresponds to a higher hydrogen ion concentration.
306. The statement “a solution with \(pH=6.8\) is always acidic” is incomplete because
ⓐ. acidity should be judged only by colour
ⓑ. every solution with \(pH\lt7\) is basic
ⓒ. neutral \(pH\) can change with temperature
ⓓ. \(pH\) is unrelated to \([\mathrm{H^+}]\)
Correct Answer: neutral \(pH\) can change with temperature
Explanation: At \(298\,\text{K}\), \(pH\lt7\) indicates an acidic solution because neutral water has \(pH=7\). However, the neutral \(pH\) depends on \(K_w\), and \(K_w\) changes with temperature. Neutrality itself means \([\mathrm{H^+}]=[\mathrm{OH^-}]\). A solution with \(pH=6.8\) could be neutral at a temperature where neutral water has that \(pH\). The temperature condition must be stated before making a final acidity classification near neutrality.
307. A table gives \(pH\) values at \(298\,\text{K}\).
| Solution | \(pH\) |
|---|
| P | \(3.0\) |
| Q | \(7.0\) |
| R | \(9.5\) |
| S | \(1.0\) |
The most basic solution is
ⓐ. Q
ⓑ. P
ⓒ. S
ⓓ. R
Correct Answer: R
Explanation: At \(298\,\text{K}\), a solution with \(pH\gt7\) is basic. Among the values given, \(9.5\) is the highest \(pH\). Higher \(pH\) means lower \([\mathrm{H^+}]\) and, at \(298\,\text{K}\), higher relative basicity. Solution \(Q\) is neutral at this temperature. Solutions \(P\) and \(S\) are acidic because their \(pH\) values are below \(7\).
308. At \(298\,\text{K}\), a solution has \([\mathrm{OH^-}]=1.0\times10^{-4}\,\text{M}\). Its \(pH\) is
ⓐ. \(10\)
ⓑ. \(14\)
ⓒ. \(4\)
ⓓ. \(7\)
Correct Answer: \(10\)
Explanation: \( \textbf{Given concentration:} \)
\[[\mathrm{OH^-}]=1.0\times10^{-4}\,\text{M}\]
\( \textbf{Find \(pOH\):} \)
\[pOH=-\log[\mathrm{OH^-}]\]
\[pOH=-\log(1.0\times10^{-4})=4\]
\( \textbf{Relation at \(298\,\text{K}\):} \)
\[pH+pOH=14\]
\( \textbf{Solve for \(pH\):} \)
\[pH=14-4=10\]
\( \textbf{Final answer:} \) \(10\). The solution is basic at \(298\,\text{K}\) because its \(pH\) is greater than \(7\).
309. The relation \(pK_a=-\log K_a\) means that, among acids at the same temperature,
ⓐ. \(pK_a\) increases when \(K_a\) increases
ⓑ. higher \(pK_a\) always indicates stronger acid
ⓒ. lower \(pK_a\) indicates stronger acid
ⓓ. \(pK_a\) is unrelated to acid strength
Correct Answer: lower \(pK_a\) indicates stronger acid
Explanation: The \(pK_a\) scale is logarithmic and has a negative sign. A larger \(K_a\) means greater acid ionization and stronger acid strength. Because \(pK_a=-\log K_a\), a larger \(K_a\) corresponds to a smaller \(pK_a\). Therefore lower \(pK_a\) means stronger acid. This is the opposite direction from comparing \(K_a\) values directly.
310. Two weak acids have \(pK_a\) values \(4.2\) and \(6.8\) at the same temperature. The stronger acid is the one with
ⓐ. \(pK_a=4.2\)
ⓑ. \(pK_a=6.8\)
ⓒ. the larger molecular mass
ⓓ. the lower concentration only
Correct Answer: \(pK_a=4.2\)
Explanation: Acid strength can be compared using \(pK_a\) when the values are measured at the same temperature. Lower \(pK_a\) means larger \(K_a\). A larger \(K_a\) means the acid ionizes to a greater extent. Therefore the acid with \(pK_a=4.2\) is stronger than the acid with \(pK_a=6.8\). Concentration and molecular mass are not the basis of this strength comparison.
311. A monoprotic strong acid of concentration \(0.010\,\text{M}\) is assumed to ionize completely. Its \(pH\) is
ⓐ. \(1\)
ⓑ. \(12\)
ⓒ. \(7\)
ⓓ. \(2\)
Correct Answer: \(2\)
Explanation: \( \textbf{Strong acid assumption:} \)
A monoprotic strong acid ionizes completely, so \([\mathrm{H^+}]\approx C\).
\( \textbf{Given concentration:} \)
\[C=0.010\,\text{M}=1.0\times10^{-2}\,\text{M}\]
\( \textbf{Hydrogen ion concentration:} \)
\[[\mathrm{H^+}]=1.0\times10^{-2}\,\text{M}\]
\( \textbf{Definition of \(pH\):} \)
\[pH=-\log[\mathrm{H^+}]\]
\( \textbf{Calculation:} \)
\[pH=-\log(1.0\times10^{-2})=2\]
\( \textbf{Final answer:} \) \(2\). The monoprotic condition matters because one mole of acid gives one mole of \(\mathrm{H^+}\).
312. A \(0.0010\,\text{M}\) solution of \(\mathrm{NaOH}\) at \(298\,\text{K}\) is treated as a strong base. Its \(pH\) is
ⓐ. \(3\)
ⓑ. \(11\)
ⓒ. \(7\)
ⓓ. \(13\)
Correct Answer: \(11\)
Explanation: \( \textbf{Strong base assumption:} \)
\(\mathrm{NaOH}\) dissociates completely, so \([\mathrm{OH^-}]\approx0.0010\,\text{M}\).
\( \textbf{Write in power form:} \)
\[[\mathrm{OH^-}]=1.0\times10^{-3}\,\text{M}\]
\( \textbf{Find \(pOH\):} \)
\[pOH=-\log(1.0\times10^{-3})=3\]
\( \textbf{Use the \(298\,\text{K}\) relation:} \)
\[pH+pOH=14\]
\( \textbf{Calculate \(pH\):} \)
\[pH=14-3=11\]
\( \textbf{Final answer:} \) \(11\). The \(pOH\) must be converted to \(pH\) for a base calculation.
313. A weak acid solution has \(K_a=1.0\times10^{-5}\) and \(C=0.10\,\text{M}\). Using \([\mathrm{H^+}]\approx\sqrt{K_aC}\), the \(pH\) is
ⓐ. \(5\)
ⓑ. \(3\)
ⓒ. \(2\)
ⓓ. \(11\)
Correct Answer: \(3\)
Explanation: \( \textbf{Approximate weak acid relation:} \)
\[[\mathrm{H^+}]\approx\sqrt{K_aC}\]
\( \textbf{Given data:} \)
\[K_a=1.0\times10^{-5},\quad C=0.10\,\text{M}=1.0\times10^{-1}\,\text{M}\]
\( \textbf{Substitution:} \)
\[[\mathrm{H^+}]\approx\sqrt{(1.0\times10^{-5})(1.0\times10^{-1})}\]
\( \textbf{Product inside square root:} \)
\[1.0\times10^{-6}\]
\( \textbf{Hydrogen ion concentration:} \)
\[[\mathrm{H^+}]\approx1.0\times10^{-3}\,\text{M}\]
\( \textbf{Calculate \(pH\):} \)
\[pH=-\log(1.0\times10^{-3})=3\]
\( \textbf{Final answer:} \) \(3\). The weak acid is not treated like a strong acid; its \([\mathrm{H^+}]\) comes from the square-root approximation.
314. A weak base solution has \(K_b=1.0\times10^{-6}\) and \(C=0.10\,\text{M}\). At \(298\,\text{K}\), its approximate \(pH\) is
ⓐ. \(5.5\)
ⓑ. \(8.5\)
ⓒ. \(3.5\) for acidity
ⓓ. \(10.5\)
Correct Answer: \(10.5\)
Explanation: \( \textbf{Approximate weak base relation:} \)
\[[\mathrm{OH^-}]\approx\sqrt{K_bC}\]
\( \textbf{Given data:} \)
\[K_b=1.0\times10^{-6},\quad C=0.10\,\text{M}=1.0\times10^{-1}\,\text{M}\]
\( \textbf{Substitution:} \)
\[[\mathrm{OH^-}]\approx\sqrt{(1.0\times10^{-6})(1.0\times10^{-1})}\]
\( \textbf{Product inside square root:} \)
\[1.0\times10^{-7}\]
\( \textbf{Square root:} \)
\[[\mathrm{OH^-}]\approx3.16\times10^{-4}\,\text{M}\]
\( \textbf{Find \(pOH\):} \)
\[pOH=-\log(3.16\times10^{-4})\approx3.5\]
\( \textbf{Convert to \(pH\) at \(298\,\text{K}\):} \)
\[pH=14-3.5=10.5\]
\( \textbf{Final answer:} \) \(10.5\). Weak base calculations usually give \(pOH\) first, so the final conversion to \(pH\) should not be skipped.
315. Adding \(\mathrm{CH_3COONa}\) to a solution of weak acid \(\mathrm{CH_3COOH}\) suppresses ionization mainly because
ⓐ. \(\mathrm{CH_3COOH}\) becomes a strong acid immediately
ⓑ. \(\mathrm{Na^+}\) removes all acid molecules from solution
ⓒ. common ion \(\mathrm{CH_3COO^-}\) shifts ionization left
ⓓ. the value of \(K_a\) increases at the same temperature
Correct Answer: common ion \(\mathrm{CH_3COO^-}\) shifts ionization left
Explanation: Acetic acid ionizes as \(\mathrm{CH_3COOH(aq)\rightleftharpoons H^+(aq)+CH_3COO^-(aq)}\). Sodium acetate supplies \(\mathrm{CH_3COO^-}\), which is already a product ion in this equilibrium. Adding a product ion shifts the equilibrium toward undissociated \(\mathrm{CH_3COOH}\). This reduces the degree of ionization of the weak acid. The value of \(K_a\) remains fixed at the same temperature; the composition changes to satisfy that same \(K_a\).
316. A weak base \(\mathrm{NH_3}\) is in equilibrium as \(\mathrm{NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)}\). Adding \(\mathrm{NH_4Cl}\) will
ⓐ. suppress ionization of \(\mathrm{NH_3}\)
ⓑ. change \(\mathrm{NH_3}\) into a strong base
ⓒ. remove all \(\mathrm{OH^-}\) permanently
ⓓ. increase ionization of \(\mathrm{NH_3}\)
Correct Answer: suppress ionization of \(\mathrm{NH_3}\)
Explanation: Ammonium chloride supplies \(\mathrm{NH_4^+}\), which is a product ion in the ionization equilibrium of \(\mathrm{NH_3}\). Adding this common ion increases the product side concentration. The equilibrium shifts left to reduce the added ion effect. This decreases the formation of \(\mathrm{OH^-}\) from \(\mathrm{NH_3}\). The weak base remains weak; the common ion changes the extent of ionization, not the identity of the base.
317. A solution contains a weak acid \(\mathrm{HA}\) and its salt \(\mathrm{NaA}\). The most direct reason this mixture can resist small changes in \(pH\) is that it contains
ⓐ. a weak acid and its conjugate base
ⓑ. a salt of strong acid and strong base only
ⓒ. only neutral molecules and no ions
ⓓ. a strong acid and a strong base in equal amounts
Correct Answer: a weak acid and its conjugate base
Explanation: A buffer solution contains a conjugate acid-base pair in appreciable amounts. In an acidic buffer, the weak acid \(\mathrm{HA}\) can neutralize added base, while the conjugate base \(\mathrm{A^-}\) can neutralize added acid. The salt \(\mathrm{NaA}\) supplies \(\mathrm{A^-}\). This pair helps keep the ratio \(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\) from changing sharply after small additions. A strong acid-strong base salt alone does not provide this reversible weak pair.
318. A mixture of \(\mathrm{NH_3}\) and \(\mathrm{NH_4Cl}\) is an example of
ⓐ. an acidic buffer containing weak acid and conjugate base
ⓑ. a neutral solution with no acid-base equilibrium
ⓒ. a strong acid-strong base mixture
ⓓ. a basic buffer containing weak base and conjugate acid
Correct Answer: a basic buffer containing weak base and conjugate acid
Explanation: \(\mathrm{NH_3}\) is a weak base. \(\mathrm{NH_4Cl}\) supplies \(\mathrm{NH_4^+}\), which is the conjugate acid of \(\mathrm{NH_3}\). A mixture containing a weak base and its conjugate acid salt behaves as a basic buffer. It can consume small amounts of added acid or base through the conjugate pair. The buffer action depends on the presence of both \(\mathrm{NH_3}\) and \(\mathrm{NH_4^+}\), not on chloride ion acting as a base.
319. The mixture least likely to act as a buffer is
ⓐ. \(\mathrm{NH_3}\) and \(\mathrm{NH_4Cl}\)
ⓑ. \(\mathrm{HCl}\) and \(\mathrm{NaCl}\)
ⓒ. \(\mathrm{H_2CO_3}\) and \(\mathrm{NaHCO_3}\)
ⓓ. \(\mathrm{CH_3COOH}\) and \(\mathrm{CH_3COONa}\)
Correct Answer: \(\mathrm{HCl}\) and \(\mathrm{NaCl}\)
Explanation: A buffer needs a weak acid with its conjugate base, or a weak base with its conjugate acid. \(\mathrm{CH_3COOH/CH_3COO^-}\), \(\mathrm{NH_3/NH_4^+}\), and \(\mathrm{H_2CO_3/HCO_3^-}\) are conjugate weak pairs. \(\mathrm{HCl}\) is a strong acid, and \(\mathrm{Cl^-}\) is the very weak conjugate base of a strong acid. The pair \(\mathrm{HCl/Cl^-}\) does not provide the reversible weak acid-base balance needed for buffer action. A salt beside a strong acid is not automatically a buffer.
320. When a small amount of strong acid is added to an acidic buffer containing \(\mathrm{HA}\) and \(\mathrm{A^-}\), the added \(\mathrm{H^+}\) is mainly consumed by
ⓐ. water to remove all buffer components
ⓑ. \(\mathrm{Na^+}\) to form metallic sodium
ⓒ. \(\mathrm{A^-}\) to form \(\mathrm{HA}\)
ⓓ. \(\mathrm{HA}\) to form \(\mathrm{A^-}\)
Correct Answer: \(\mathrm{A^-}\) to form \(\mathrm{HA}\)
Explanation: In an acidic buffer, \(\mathrm{A^-}\) is the conjugate base component. When a small amount of strong acid is added, it supplies \(\mathrm{H^+}\). The conjugate base \(\mathrm{A^-}\) accepts this proton and forms \(\mathrm{HA}\). This reaction reduces the sudden increase in \([\mathrm{H^+}]\). Buffer action is strongest when both \(\mathrm{HA}\) and \(\mathrm{A^-}\) are present in appreciable amounts.
