91Ó°ÊÓ

(a) The conjugate acid of ${\text{H}}_2\text{O}$is ${\text{H}}_3{\text{O}}^{+}$with the ${\text{pK}}_{\text{a}}$value of -1.7. Since the ${\text{pK}}_{\text{a}}$ofrole="math" localid="1649243149644" ${\text{H}}_3{\text{O}}^{+}$ is lower than the ${\text{pK}}_{\text{a}}$of ${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{°ä≡C±á}$, so, ${\text{H}}_2\text{O}$cannot deprotonate the given acid.

(b) The conjugate acid of NaOH is${\text{H}}_2\text{O}$ with the ${\text{pK}}_{\text{a}}$value of 15.7. Since the ${\text{pK}}_{\text{a}}$of${\text{H}}_2\text{O}$ is lower than the ${\text{pK}}_{\text{a}}$of ${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{°ä≡C±á}$, so, NaOH cannot deprotonate the given acid.

(c) The conjugate acid ${\text{NaNH}}_2$of ${\text{NH}}_3$is with the ${\text{pK}}_{\text{a}}$value of 38. Since the ${\text{pK}}_{\text{a}}$of${\text{NH}}_3$ is higher than the ${\text{pK}}_{\text{a}}$of ${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{°ä≡C±á}$, NaOH is strong enough as a base to deprotonate the given acid.

(d) The conjugate acid of${\text{NH}}_3$ is${\text{NH}}_4^{+}$ with the ${\text{pK}}_{\text{a}}$value of 9.4. Since the ${\text{pK}}_{\text{a}}$of${\text{NH}}_4^{+}$ is lower than the ${\text{pK}}_{\text{a}}$of ${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{°ä≡C±á}$, so, ${\text{NH}}_3$cannot deprotonate the given acid.

(e) The conjugate acid of NaH is${\text{H}}_2$ with the ${\text{pK}}_{\text{a}}$value of 35. Since the ${\text{pK}}_{\text{a}}$of${\text{H}}_2$ is higher than the ${\text{pK}}_{\text{a}}$of ${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{°ä≡C±á}$, NaH is strong enough as a base to deprotonate the given acid.

(f) The conjugate acid of ${\text{CH}}_3\text{Li}$ is${\text{CH}}_4$ with the ${\text{pK}}_{\text{a}}$value of 50. Since the ${\text{pK}}_{\text{a}}$of${\text{CH}}_4$ is higher than the ${\text{pK}}_{\text{a}}$of ${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{°ä≡C±á}$, ${\text{CH}}_3\text{Li}$is strong enough as a base to deprotonate the given acid.