401. A passage record says: A closed vessel contains \(\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}\) at equilibrium. The vessel is compressed at constant temperature, and later a catalyst is introduced. The correct combined prediction is
ⓐ. compression shifts left; catalyst only speeds re-equilibration
ⓑ. compression shifts right, and the catalyst increases \(K_p\)
ⓒ. compression has no effect, and the catalyst removes the reverse reaction
ⓓ. compression shifts left, and the catalyst changes the final equilibrium composition
Correct Answer: compression shifts left; catalyst only speeds re-equilibration
Explanation: In \(\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}\), the left side has \(1\) gaseous mole and the right side has \(2\) gaseous moles. Compression increases pressure, so the equilibrium shifts toward the side with fewer gaseous moles. That direction is left, toward \(\mathrm{PCl_5}\). A catalyst affects the rate of both forward and reverse reactions. It does not change \(K_p\) or the final equilibrium composition at the same temperature.
402. A graph description for an exothermic reaction \(\mathrm{A\rightleftharpoons B}\) shows that after heating, the concentration of \(\mathrm{B}\) gradually decreases to a new plateau. The forward reaction is most likely
ⓐ. unaffected by temperature
ⓑ. irreversible after heating
ⓒ. endothermic with \(\mathrm{B}\) as product
ⓓ. exothermic with \(\mathrm{B}\) as product
Correct Answer: exothermic with \(\mathrm{B}\) as product
Explanation: Heating favours the endothermic direction of a reversible reaction. If \(\mathrm{B}\) is the product and its concentration decreases after heating, the equilibrium has shifted in the reverse direction. That means the forward direction, which forms \(\mathrm{B}\), is disfavoured by heating. A direction disfavoured by heating is exothermic. The new plateau shows a new equilibrium composition, not an irreversible stopping point.
403. A reaction has \(K_c=25\) for \(\mathrm{A\rightleftharpoons B}\). A current mixture has \([\mathrm{A}]=0.020\,\text{M}\) and \([\mathrm{B}]=0.50\,\text{M}\). The mixture is
ⓐ. shifting forward
ⓑ. at equilibrium
ⓒ. impossible because \([\mathrm{B}]\gt[\mathrm{A}]\)
ⓓ. shifting in reverse
Correct Answer: at equilibrium
Explanation: \( \textbf{Reaction quotient:} \)
\[Q_c=\frac{[\mathrm{B}]}{[\mathrm{A}]}\]
\( \textbf{Current concentrations:} \)
\[[\mathrm{B}]=0.50\,\text{M},\quad [\mathrm{A}]=0.020\,\text{M}\]
\( \textbf{Calculate \(Q_c\):} \)
\[Q_c=\frac{0.50}{0.020}=25\]
\( \textbf{Compare with \(K_c\):} \)
\[Q_c=25,\quad K_c=25\]
\( \textbf{Conclusion:} \)
\[Q_c=K_c\]
\( \textbf{Final answer:} \) at equilibrium. A product-rich mixture can be exactly at equilibrium when the equilibrium constant is also large.
404. A reaction has \(\Delta G^\circ=+11.4\,\text{kJ mol}^{-1}\) at \(298\,\text{K}\). Which interpretation is most consistent with \( \Delta G^\circ=-RT\ln K \)?
ⓐ. \(K=1\), so neither side is favoured
ⓑ. \(K\) is unrelated to \(\Delta G^\circ\)
ⓒ. \(K\lt1\), so reactants are favoured
ⓓ. \(K\gt1\), so products are strongly favoured
Correct Answer: \(K\lt1\), so reactants are favoured
Explanation: The relation is \(\Delta G^\circ=-RT\ln K\). Since \(R\) and \(T\) are positive, the sign of \(\Delta G^\circ\) is opposite to the sign of \(\ln K\). A positive \(\Delta G^\circ\) means \(\ln K\) is negative. A negative \(\ln K\) corresponds to \(K\lt1\). Therefore the written reaction is reactant-favoured under standard-state comparison.
405. A salt solution is made from a cation that is the conjugate acid of a weak base and an anion that is the conjugate base of a strong acid. The solution is expected to be
ⓐ. acidic
ⓑ. basic only if the anion hydrolyses strongly
ⓒ. neutral because all salts are neutral
ⓓ. basic
Correct Answer: acidic
Explanation: A cation that is the conjugate acid of a weak base can donate a proton to water. This hydrolysis produces \(\mathrm{H_3O^+}\) and makes the solution acidic. The anion from a strong acid is usually too weak as a base to hydrolyse appreciably. Therefore the cation hydrolysis dominates the solution nature. This is the pattern for salts such as \(\mathrm{NH_4Cl}\).
406. A salt solution is made from an anion that is the conjugate base of a weak acid and a cation from a strong base. The hydrolysis reaction most directly responsible for basicity is
ⓐ. \(\mathrm{HA(aq)\rightleftharpoons H_2(g)+A(s)}\)
ⓑ. \(\mathrm{M^+(aq)+H_2O(l)\rightleftharpoons MOH(s)+H^+(aq)}\)
ⓒ. \(\mathrm{A^-(aq)+H_2O(l)\rightleftharpoons HA(aq)+OH^-(aq)}\)
ⓓ. \(\mathrm{A^-(aq)+H_2O(l)\rightleftharpoons AOH(aq)+H^+(aq)}\)
Correct Answer: \(\mathrm{A^-(aq)+H_2O(l)\rightleftharpoons HA(aq)+OH^-(aq)}\)
Explanation: The anion \(\mathrm{A^-}\) is the conjugate base of a weak acid \(\mathrm{HA}\). It can accept a proton from water. This produces undissociated \(\mathrm{HA}\) and \(\mathrm{OH^-}\). The formation of \(\mathrm{OH^-}\) makes the solution basic. The cation from a strong base usually behaves as a spectator in this hydrolysis classification.
407. A sparingly soluble salt \(\mathrm{AB_2}\) has \(K_{sp}=3.2\times10^{-8}\). For \(\mathrm{AB_2(s)\rightleftharpoons A^{2+}(aq)+2B^-(aq)}\), the molar solubility is
ⓐ. \(4.0\times10^{-3}\,\text{M}\)
ⓑ. \(1.0\times10^{-3}\,\text{M}\)
ⓒ. \(8.0\times10^{-3}\,\text{M}\)
ⓓ. \(2.0\times10^{-3}\,\text{M}\)
Correct Answer: \(2.0\times10^{-3}\,\text{M}\)
Explanation: \( \textbf{Salt pattern:} \)
\[\mathrm{AB_2(s)\rightleftharpoons A^{2+}(aq)+2B^-(aq)}\]
\( \textbf{Ion concentrations from molar solubility \(s\):} \)
\[[\mathrm{A^{2+}}]=s,\quad [\mathrm{B^-}]=2s\]
\( \textbf{Solubility product:} \)
\[K_{sp}=[\mathrm{A^{2+}}][\mathrm{B^-}]^2=s(2s)^2=4s^3\]
\( \textbf{Substitution:} \)
\[3.2\times10^{-8}=4s^3\]
\( \textbf{Solve for \(s^3\):} \)
\[s^3=8.0\times10^{-9}\]
\( \textbf{Cube root:} \)
\[s=2.0\times10^{-3}\,\text{M}\]
\( \textbf{Final answer:} \) \(2.0\times10^{-3}\,\text{M}\). The factor \(4\) comes from the two \(B^-\) ions produced per formula unit.
408. A vessel contains \(\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}\). At equilibrium, \(p_{\mathrm{N_2O_4}}=0.40\,\text{atm}\) and \(p_{\mathrm{NO_2}}=0.20\,\text{atm}\). The value of \(K_p\) is
ⓐ. \(0.40\,\text{atm}\)
ⓑ. \(0.10\,\text{atm}\)
ⓒ. \(0.20\,\text{atm}\)
ⓓ. \(2.0\,\text{atm}\)
Correct Answer: \(0.10\,\text{atm}\)
Explanation: \( \textbf{Equilibrium reaction:} \)
\[\mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)}\]
\( \textbf{Pressure expression:} \)
\[K_p=\frac{(p_{\mathrm{NO_2}})^2}{p_{\mathrm{N_2O_4}}}\]
\( \textbf{Given partial pressures:} \)
\[p_{\mathrm{NO_2}}=0.20\,\text{atm},\quad p_{\mathrm{N_2O_4}}=0.40\,\text{atm}\]
\( \textbf{Substitution:} \)
\[K_p=\frac{(0.20)^2}{0.40}\]
\( \textbf{Calculate numerator:} \)
\[(0.20)^2=0.040\]
\( \textbf{Final calculation:} \)
\[K_p=\frac{0.040}{0.40}=0.10\,\text{atm}\]
\( \textbf{Final answer:} \) \(0.10\,\text{atm}\). The pressure of \(\mathrm{NO_2}\) must be squared because its coefficient is \(2\).
409. A reaction \(\mathrm{A(g)+B(g)\rightleftharpoons2C(g)}\) has \(K_c=9.0\). Initially, \([\mathrm{A}]=[\mathrm{B}]=1.00\,\text{M}\) and \([\mathrm{C}]=0\). If \(x\) is the amount of \(\mathrm{A}\) consumed, the equation for \(x\) is
ⓐ. \(9.0=\frac{(1.00-x)^2}{(2x)^2}\)
ⓑ. \(9.0=\frac{(2x)^2}{(1.00-x)^2}\)
ⓒ. \(9.0=\frac{x^2}{1.00-x}\)
ⓓ. \(9.0=\frac{x}{(1.00-x)^2}\)
Correct Answer: \(9.0=\frac{(2x)^2}{(1.00-x)^2}\)
Explanation: The balanced reaction is \(\mathrm{A(g)+B(g)\rightleftharpoons2C(g)}\). If \(x\) of \(\mathrm{A}\) is consumed, \(x\) of \(\mathrm{B}\) is also consumed. The concentration of \(\mathrm{C}\) formed is \(2x\) because the product coefficient is \(2\). Therefore \([\mathrm{A}]_{\text{eq}}=1.00-x\), \([\mathrm{B}]_{\text{eq}}=1.00-x\), and \([\mathrm{C}]_{\text{eq}}=2x\). Substitution into \(K_c=\frac{[\mathrm{C}]^2}{[\mathrm{A}][\mathrm{B}]}\) gives \(9.0=\frac{(2x)^2}{(1.00-x)^2}\). The factor \(2\) must stay inside the square.
410. For \(\mathrm{A(g)+B(g)\rightleftharpoons2C(g)}\), a current mixture has \(p_{\mathrm{A}}=0.20\,\text{bar}\), \(p_{\mathrm{B}}=0.50\,\text{bar}\), and \(p_{\mathrm{C}}=0.40\,\text{bar}\). If \(K_p=2.0\), the initial shift is
ⓐ. impossible because pressure values are unequal
ⓑ. reverse
ⓒ. no net shift
ⓓ. forward
Correct Answer: forward
Explanation: \( \textbf{Pressure quotient expression:} \)
\[Q_p=\frac{(p_{\mathrm{C}})^2}{p_{\mathrm{A}}p_{\mathrm{B}}}\]
\( \textbf{Given current partial pressures:} \)
\[p_{\mathrm{A}}=0.20\,\text{bar},\quad p_{\mathrm{B}}=0.50\,\text{bar},\quad p_{\mathrm{C}}=0.40\,\text{bar}\]
\( \textbf{Calculate \(Q_p\):} \)
\[Q_p=\frac{(0.40)^2}{(0.20)(0.50)}\]
\[Q_p=\frac{0.16}{0.10}=1.6\]
\( \textbf{Compare with \(K_p\):} \)
\[Q_p=1.6,\quad K_p=2.0\]
\[Q_p\lt K_p\]
\( \textbf{Final answer:} \) forward. The mixture needs a larger product pressure term to reach the equilibrium value.
411. A weak acid \(\mathrm{HA}\) has \(K_a=1.0\times10^{-5}\), and a sparingly soluble salt \(\mathrm{MA}\) has \(K_{sp}=1.0\times10^{-10}\). Adding strong acid to a saturated \(\mathrm{MA}\) suspension containing \(\mathrm{A^-}\) tends to
ⓐ. lower \([\mathrm{A^-}]\), so more \(\mathrm{MA}\) dissolves
ⓑ. increase \([\mathrm{A^-}]\) by forming more salt immediately
ⓒ. make \(K_{sp}\) larger without changing any concentration
ⓓ. stop all acid-base reactions because a solid is present
Correct Answer: lower \([\mathrm{A^-}]\), so more \(\mathrm{MA}\) dissolves
Explanation: The anion \(\mathrm{A^-}\) can accept added \(\mathrm{H^+}\) to form the weak acid \(\mathrm{HA}\). This removes some free \(\mathrm{A^-}\) from solution. In the solubility equilibrium \(\mathrm{MA(s)\rightleftharpoons M^+(aq)+A^-(aq)}\), lowering a product ion concentration shifts dissolution to the right. More \(\mathrm{MA}\) can dissolve to restore the solubility-product condition. This is an integrated common-ion and acid-base effect, not a change in \(K_{sp}\) at fixed temperature.
412. A learner writes \(K_{sp}=s^2\) for \(\mathrm{CaF_2}\), \(\mathrm{AgCl}\), and \(\mathrm{Al_2S_3}\). The best correction is that
ⓐ. \(K_{sp}\) never depends on stoichiometric coefficients
ⓑ. \(K_{sp}=s^2\) applies to every sparingly soluble salt
ⓒ. \(K_{sp}=s^2\) applies only to \(1:1\) salts
ⓓ. \(s\) is not related to ion concentration
Correct Answer: \(K_{sp}=s^2\) applies only to \(1:1\) salts
Explanation: The relation between \(K_{sp}\) and molar solubility depends on the salt formula. For \(\mathrm{AgCl}\), \([\mathrm{Ag^+}]=s\) and \([\mathrm{Cl^-}]=s\), so \(K_{sp}=s^2\). For \(\mathrm{CaF_2}\), \([\mathrm{Ca^{2+}}]=s\) and \([\mathrm{F^-}]=2s\), so \(K_{sp}=4s^3\). For \(\mathrm{Al_2S_3}\), the relation becomes \(K_{sp}=108s^5\). Stoichiometric coefficients must be converted into ion concentrations before writing the final \(s\)-relation.
413. In a mixed equilibrium problem, the solution contains a weak base \(\mathrm{B}\), its conjugate acid \(\mathrm{BH^+}\), and a salt whose cation hydrolyses. The most reliable way to begin is to
ⓐ. assume \(\mathrm{BH^+}\) is always neutral
ⓑ. choose the relevant equilibrium constant
ⓒ. use \(pH=-\log[\mathrm{B}]\) for any weak base
ⓓ. use \(K_{sp}\) for every ion in the solution
Correct Answer: choose the relevant equilibrium constant
Explanation: Mixed ionic equilibrium problems often contain several related species. A weak base and its conjugate acid suggest a buffer relation. A hydrolysing cation may require \(K_h=\frac{K_w}{K_b}\) if it comes from a weak base. A sparingly soluble salt would require \(K_{sp}\) and possibly \(Q_{sp}\). Choosing the equation before identifying the equilibrium type can give a numerically neat but chemically wrong answer. Species identity, charge, and the process being asked about should decide the formula.
414. For \(\mathrm{AB_3(s)\rightleftharpoons A^{3+}(aq)+3B^-(aq)}\), if the molar solubility is \(s\), the correct relation between \(K_{sp}\) and \(s\) is
ⓐ. \(K_{sp}=s^2\)
ⓑ. \(K_{sp}=3s^2\)
ⓒ. \(K_{sp}=27s^4\)
ⓓ. \(K_{sp}=81s^4\)
Correct Answer: \(K_{sp}=27s^4\)
Explanation: \( \textbf{Dissolution equation:} \)
\[\mathrm{AB_3(s)\rightleftharpoons A^{3+}(aq)+3B^-(aq)}\]
\( \textbf{Ion concentrations from molar solubility:} \)
\[[\mathrm{A^{3+}}]=s\]
\[[\mathrm{B^-}]=3s\]
\( \textbf{Solubility product expression:} \)
\[K_{sp}=[\mathrm{A^{3+}}][\mathrm{B^-}]^3\]
\( \textbf{Substitution:} \)
\[K_{sp}=s(3s)^3\]
\( \textbf{Simplification:} \)
\[(3s)^3=27s^3\]
\[K_{sp}=s(27s^3)=27s^4\]
\( \textbf{Final answer:} \) \(27s^4\). The coefficient \(3\) affects both the ion concentration and the exponent in the \(K_{sp}\) expression.
415. Equal volumes of \(1.0\times10^{-3}\,\text{M}\) \(\mathrm{Pb(NO_3)_2}\) and \(2.0\times10^{-3}\,\text{M}\) \(\mathrm{KI}\) are mixed. If \(K_{sp}(\mathrm{PbI_2})=8.0\times10^{-9}\), the correct prediction is
ⓐ. no precipitate forms because \(Q_{sp}\lt K_{sp}\)
ⓑ. \(\mathrm{PbI_2}\) precipitates because \(Q_{sp}\gt K_{sp}\)
ⓒ. the mixture is exactly saturated because the ion concentrations are equal
ⓓ. no precipitate forms because both original salts are soluble
Correct Answer: no precipitate forms because \(Q_{sp}\lt K_{sp}\)
Explanation: \( \textbf{Dissolution expression for \(\mathrm{PbI_2}\):} \)
\[\mathrm{PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)}\]
\( \textbf{Concentrations after mixing equal volumes:} \)
\[[\mathrm{Pb^{2+}}]=\frac{1.0\times10^{-3}}{2}=5.0\times10^{-4}\,\text{M}\]
\[[\mathrm{I^-}]=\frac{2.0\times10^{-3}}{2}=1.0\times10^{-3}\,\text{M}\]
\( \textbf{Ionic product:} \)
\[Q_{sp}=[\mathrm{Pb^{2+}}][\mathrm{I^-}]^2\]
\( \textbf{Substitution:} \)
\[Q_{sp}=(5.0\times10^{-4})(1.0\times10^{-3})^2\]
\( \textbf{Calculation:} \)
\[(1.0\times10^{-3})^2=1.0\times10^{-6}\]
\[Q_{sp}=5.0\times10^{-10}\]
\( \textbf{Comparison:} \)
\[5.0\times10^{-10}\lt8.0\times10^{-9}\]
Since \(Q_{sp}\lt K_{sp}\), the mixture is unsaturated with respect to \(\mathrm{PbI_2}\). \( \textbf{Final answer:} \) no precipitate forms because \(Q_{sp}\lt K_{sp}\).
416. Equal volumes of \(1.0\times10^{-2}\,\text{M}\) \(\mathrm{Pb(NO_3)_2}\) and \(2.0\times10^{-2}\,\text{M}\) \(\mathrm{KI}\) are mixed. If \(K_{sp}(\mathrm{PbI_2})=8.0\times10^{-9}\), the correct prediction is
ⓐ. precipitation is impossible because nitrate salts are soluble
ⓑ. no precipitate forms because \(Q_{sp}\lt K_{sp}\)
ⓒ. the solution is exactly saturated because the volumes are equal
ⓓ. \(\mathrm{PbI_2}\) precipitates because \(Q_{sp}\gt K_{sp}\)
Correct Answer: \(\mathrm{PbI_2}\) precipitates because \(Q_{sp}\gt K_{sp}\)
Explanation: \( \textbf{Dissolution equilibrium:} \)
\[\mathrm{PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)}\]
\( \textbf{Ion concentrations after equal-volume mixing:} \)
\[[\mathrm{Pb^{2+}}]=5.0\times10^{-3}\,\text{M}\]
\[[\mathrm{I^-}]=1.0\times10^{-2}\,\text{M}\]
\( \textbf{Ionic product expression:} \)
\[Q_{sp}=[\mathrm{Pb^{2+}}][\mathrm{I^-}]^2\]
\( \textbf{Substitution:} \)
\[Q_{sp}=(5.0\times10^{-3})(1.0\times10^{-2})^2\]
\( \textbf{Square iodide concentration:} \)
\[(1.0\times10^{-2})^2=1.0\times10^{-4}\]
\( \textbf{Calculate \(Q_{sp}\):} \)
\[Q_{sp}=5.0\times10^{-7}\]
\( \textbf{Compare with \(K_{sp}\):} \)
\[5.0\times10^{-7}\gt 8.0\times10^{-9}\]
\( \textbf{Final answer:} \) \(\mathrm{PbI_2}\) precipitates. Equal-volume dilution must be applied before comparing \(Q_{sp}\) with \(K_{sp}\).
417. In a \(0.010\,\text{M}\) \(\mathrm{NaOH}\) solution, the molar solubility of \(\mathrm{Mg(OH)_2}\) is calculated using \(K_{sp}=1.6\times10^{-11}\). The approximate solubility is
ⓐ. \(1.6\times10^{-5}\,\text{M}\)
ⓑ. \(4.0\times10^{-6}\,\text{M}\)
ⓒ. \(1.6\times10^{-7}\,\text{M}\)
ⓓ. \(1.6\times10^{-9}\,\text{M}\)
Correct Answer: \(1.6\times10^{-7}\,\text{M}\)
Explanation: \( \textbf{Dissolution equilibrium:} \)
\[\mathrm{Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)}\]
\( \textbf{Solubility product:} \)
\[K_{sp}=[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2\]
\( \textbf{Common ion concentration:} \)
\[[\mathrm{OH^-}]\approx0.010\,\text{M}=1.0\times10^{-2}\,\text{M}\]
\( \textbf{Let solubility be \(s\):} \)
\[[\mathrm{Mg^{2+}}]=s\]
\( \textbf{Substitution:} \)
\[1.6\times10^{-11}=s(1.0\times10^{-2})^2\]
\( \textbf{Square hydroxide concentration:} \)
\[(1.0\times10^{-2})^2=1.0\times10^{-4}\]
\( \textbf{Solve for \(s\):} \)
\[s=\frac{1.6\times10^{-11}}{1.0\times10^{-4}}=1.6\times10^{-7}\,\text{M}\]
\( \textbf{Final answer:} \) \(1.6\times10^{-7}\,\text{M}\). The common hydroxide ion greatly suppresses the solubility of \(\mathrm{Mg(OH)_2}\).
418. A metal hydroxide \(\mathrm{M(OH)_2}\) has \(K_{sp}=4.0\times10^{-12}\). If \([\mathrm{M^{2+}}]=1.0\times10^{-3}\,\text{M}\), the \([\mathrm{OH^-}]\) needed to just begin precipitation is
ⓐ. \(2.0\times10^{-9}\,\text{M}\)
ⓑ. \(4.0\times10^{-9}\,\text{M}\)
ⓒ. \(4.0\times10^{-3}\,\text{M}\)
ⓓ. \(2.0\times10^{-5}\,\text{M}\)
Correct Answer: \(2.0\times10^{-5}\,\text{M}\)
Explanation: \( \textbf{Precipitation condition:} \)
\[Q_{sp}=K_{sp}\]
\( \textbf{For \(\mathrm{M(OH)_2}\):} \)
\[K_{sp}=[\mathrm{M^{2+}}][\mathrm{OH^-}]^2\]
\( \textbf{Given values:} \)
\[K_{sp}=4.0\times10^{-12},\quad [\mathrm{M^{2+}}]=1.0\times10^{-3}\,\text{M}\]
\( \textbf{Solve for hydroxide concentration:} \)
\[[\mathrm{OH^-}]^2=\frac{4.0\times10^{-12}}{1.0\times10^{-3}}\]
\[[\mathrm{OH^-}]^2=4.0\times10^{-9}\]
\( \textbf{Square root:} \)
\[[\mathrm{OH^-}]=2.0\times10^{-5}\,\text{M}\]
\( \textbf{Final answer:} \) \(2.0\times10^{-5}\,\text{M}\). The hydroxide concentration is squared because two hydroxide ions appear in the salt formula.
419. A metal hydroxide starts precipitating when \([\mathrm{OH^-}]=1.0\times10^{-6}\,\text{M}\). At \(298\,\text{K}\), the corresponding \(pH\) is
ⓐ. \(10\)
ⓑ. \(8\)
ⓒ. \(12\)
ⓓ. \(6\)
Correct Answer: \(8\)
Explanation: \( \textbf{Given hydroxide concentration:} \)
\[[\mathrm{OH^-}]=1.0\times10^{-6}\,\text{M}\]
\( \textbf{Find \(pOH\):} \)
\[pOH=-\log[\mathrm{OH^-}]\]
\[pOH=-\log(1.0\times10^{-6})=6\]
\( \textbf{Use the \(298\,\text{K}\) relation:} \)
\[pH+pOH=14\]
\( \textbf{Calculate \(pH\):} \)
\[pH=14-6=8\]
\( \textbf{Final answer:} \) \(8\). Precipitation controlled by hydroxide concentration can be converted into a \(pH\) condition at a specified temperature.
420. A \(1.00\,\text{L}\) acidic buffer contains \(0.300\,\text{mol}\) of \(\mathrm{HA}\) and \(0.200\,\text{mol}\) of \(\mathrm{A^-}\). If \(0.050\,\text{mol}\) of \(\mathrm{NaOH}\) is added and volume change is neglected, the new ratio \(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\) is
ⓐ. \(\frac{0.250}{0.250}\)
ⓑ. \(\frac{0.200}{0.300}\)
ⓒ. \(\frac{0.350}{0.150}\)
ⓓ. \(\frac{0.150}{0.350}\)
Correct Answer: \(\frac{0.250}{0.250}\)
Explanation: \( \textbf{Buffer reaction with added base:} \)
\[\mathrm{HA+OH^-\to A^-+H_2O}\]
\( \textbf{Initial moles:} \)
\[\mathrm{HA}=0.300\,\text{mol},\quad \mathrm{A^-}=0.200\,\text{mol}\]
\( \textbf{Added strong base:} \)
\[0.050\,\text{mol}\ \mathrm{OH^-}\]
\( \textbf{Mole changes:} \)
\[\mathrm{HA}:0.300-0.050=0.250\,\text{mol}\]
\[\mathrm{A^-}:0.200+0.050=0.250\,\text{mol}\]
\( \textbf{New buffer ratio:} \)
\[\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}=\frac{0.250}{0.250}\]
\( \textbf{Final answer:} \) \(\frac{0.250}{0.250}\). Added base consumes weak acid and forms conjugate base in the same mole amount.
