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Chapter 16: Problem 85

Explain why \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in a strong acid solution doubles as the total acid concentration doubles, whereas in a weak acid solution, \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) increases only by about a factor of \(\sqrt{2}\)

Short Answer

Expert verified
The concentration of \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) ion doubles in a strong acid solution when the total acid concentration doubles, because strong acids fully ionize, releasing equal amount of \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) ions. In the case of weak acids, they only partially ionize, and hence when the total acid concentration doubles, the \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) ion concentration only increases by \(\sqrt{2}\) due to the square root dependence relationship.

Step by step solution

01

Understanding Strong Acids

Strong acids fully ionize in solution. Hence, if the concentration of strong acid in the solution doubles, the concentration of \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) also doubles. This is because the number of free \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) ions is exactly equal to the number of acid molecules introduced.
02

Understanding Weak Acids

Weak acids partially ionize in solution. Thus, even if the concentration of the weak acid doubles, it does not mean that the concentration of \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) will also double. In this case, it only increases by a factor of \(\sqrt{2}\). This is due to the square root dependence relationship between \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) concentration and the initial weak acid concentration, derived from the equilibrium expression for weak acid ionisation.
03

Conclusion

Therefore, in a strong acid solution, the concentration of the \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) ion doubles as the total acid concentration doubles, because of complete ionisation. Conversely, for a weak acid solution, even with doubling the acid concentration, the concentration of the \(\[\mathrm{H}_{3} \mathrm{O}^{+}\]\) ion only increases by about a factor of \(\sqrt{2}\), due to partial ionisation.

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Most popular questions from this chapter

Maleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats. In a combustion analysis, a 1.054 g sample of maleic acid yields \(1.599 \mathrm{g}\) \(\mathrm{CO}_{2}\) and \(0.327 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In a freezing-point depression experiment, a \(0.615 \mathrm{g}\) sample of maleic acid dissolved in 25.10 g of glacial acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}(1) \quad\) (which has the freezing-point depression constant \(K_{\mathrm{f}}=3.90^{\circ} \mathrm{C} m^{-1}\) and in which maleic acid does not ionize), lowers the freezing point by \(0.82^{\circ} \mathrm{C} .\) In a titration experiment, a \(0.4250 \mathrm{g}\) sample of maleic acid is dissolved in water and requires \(34.03 \mathrm{mL}\) of \(0.2152 \mathrm{M} \mathrm{KOH}\) for its complete neutralization. The \(\mathrm{pH}\) of a \(0.215 \mathrm{g}\) sample of maleic acid dissolved in \(50.00 \mathrm{mL}\) of aqueous solution is found to be \(1.80 .\) (a) Determine the empirical and molecular formulas of maleic acid. [Hint: Which experiment(s) provide the necessary data?] (b) Use the results of part (a) and the titration data to rewrite the molecular formula to reflect the number of ionizable \(\mathrm{H}\) atoms in the molecule. (c) Given that the ionizable \(\mathrm{H}\) atom(s) is(are) associated with the carboxyl group(s), write the plausible condensed structural formula of maleic acid. (d) Determine the ionization constant(s) of maleic acid. If the data supplied are insufficient, indicate what additional data would be needed. (e) Calculate the expected \(\mathrm{pH}\) of a \(0.0500 \mathrm{M}\) aqueous solution of maleic acid. Indicate any assumptions required in this calculation.

In \(0.10 \mathrm{M} \quad \mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq}), \quad\) (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.10 \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=0.10 \mathrm{M} ;(\mathrm{c}) \mathrm{pH}<7 ;(\mathrm{d}) \mathrm{pH}<13\).

The equilibria \(\mathrm{OH}^{-}+\mathrm{HClO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{ClO}^{-}\) and \(\mathrm{ClO}^{-}+\mathrm{HNO}_{2} \longrightarrow \mathrm{HClO}+\mathrm{NO}_{2}^{-}\) both lie to the right. Which of the following is a list of acids ranked in order of decreasing strength? (a) \(\mathrm{HClO}>\mathrm{HNO}_{2}>\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{ClO}^{-}>\mathrm{NO}_{2}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{NO}_{2}^{-}>\mathrm{ClO}^{-}>\mathrm{OH}\) (d) \(\mathrm{HNO}_{2}>\mathrm{HClO}>\mathrm{H}_{2} \mathrm{O}\) (e) none of these

The conjugate acid of \(\mathrm{HPO}_{4}^{2-}\) is (a) \(\mathrm{PO}_{4}^{3-}\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} ;(\mathrm{c}) \mathrm{H}_{3} \mathrm{PO}_{4} ;(\mathrm{d}) \mathrm{H}_{3} \mathrm{O}^{+} ;\) (e) none of these.

It is possible to write simple equations to relate \(\mathrm{pH}\) \(\mathrm{p} K,\) and molarities \((\mathrm{M})\) of various solutions. Three such equations are shown here. $$\begin{aligned} &\text {Weak acid: } \quad \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{M}\\\ &\text { Weak base: } \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}+\frac{1}{2} \log \mathrm{M} \end{aligned}$$ Salt ofweak \(\operatorname{acid}\left(\mathrm{pK}_{\mathrm{a}}\right)\) and strong base: \(\quad \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{p} K_{\mathrm{a}}+\frac{1}{2} \log \mathrm{M}\) (a) Derive these three equations, and point out the assumptions involved in the derivations. (b) Use these equations to determine the pH of 0.10 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}), 0.10 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}),\) and \(0.10 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} .\) Verify that the equations give correct results by determining these pH values in the usual way.
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