Question
Calculate the $\mathrm{pH}$ of a $0.200-\mathrm{M}$ solution of $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}$. Hint: $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}$ is a salt composed of $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$ and $\mathrm{F}^{-}$ ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is$$\begin{aligned}\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F} &(a q) \rightleftharpoons \\\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$
Step 1
The reaction forming the weak acid HF has a very small K value, so very little HF is going to be in solution as an acid. Therefore, the predominant acid is going to be $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$. Show more…
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Calculate the $\mathrm{pH}$ of a $0.200-M$ solution of $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}$. Hint: $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}$ is a salt composed of $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}$ and $\mathrm{F}^{-}$ ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$
Calculate the $\mathrm{pH}$ at $25^{\circ} \mathrm{C}$ of a solution that is $0.10 \mathrm{M}$ in $\mathrm{TlBr}_{\mathrm{s}}(a q)$. The acid-dissociation constant at $25^{\circ} \mathrm{C}$ for the equilibrium described by $$ \begin{aligned} \left[\mathrm{Tl}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(a q)+& \mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \\ & \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\left[\mathrm{Tl}(\mathrm{OH})\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right]^{2+}(a q) \end{aligned} $$ is $K_{\mathrm{a}}=7.0 \times 10^{-2} \mathrm{M}$
Equilibrium constant for the following reaction is $1 \times 10^{-9}$ : $$ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(\text { aq. })+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\left(a q_{.}\right)+\mathrm{OH}^{-}(a q) $$ Determine the mole of pyridinium chloride $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} . \mathrm{HCl}\right)$ that should be added to $500 \mathrm{~mL}$ solution of $0.4 M$ pyridine $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)$ to obtain a buffer solution of $\mathrm{pH}=5$ : (a) $0.1$ mole (b) $0.2$ mole (c) $0.3$ mole (d) $0.4$ mole
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