Tutor profile: Corban E.

For a buffer of $$HX(aq)$$ and $$NaX(aq)$$, what is $$[HX]$$ if $$[X^-]=2$$ and $$pH=3$$? For $$HX\ K_a=1.3\times 10^{-3}$$.

Use the henderson equation. $$pH=pK_{a}+\log \left(\frac{[\text{conj.base}]}{[\text{weak acid }]}\right)$$ $$HX$$ is the weak acid. $$X^-$$ is the conjugate base. Substitute and solve. $$3=-log(1.3\times 10^{-3})+log\left(\frac{2}{x}\right)$$ $$x=[X^-]=1.5\ M$$

Which weak acid solution has a greater percent ionization? $$0.10\ M\ HX(aq)\ pK_a=2$$ $$0.10\ M\ HZ(aq)\ pK_a=3$$

For HX, $$K_a=10^{-2}=0.01$$. For HZ, $$K_a=10^{-3}=0.001$$. Since HX has a greater $K_a$, it has a more product favored dissociation into its ions, and therefore has a greater percent ionization. This is consistent with it being a stronger acid with a larger $$K_a$$.

If $$\Delta H^{\circ}<0$$ and $$\Delta S^{\circ}>0$$, determine what can be concluded about the following: $$K_{eq}$$ $$\Delta G^{\circ}$$ $$E^{\circ}_{cell}$$

Since $$\Delta H^{\circ}<0$$ and $$\Delta S^{\circ}>0$$, $$\Delta G^{\circ}<0$$ at all temperatures by the equation $$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$$. By the equation $$\Delta G^{\circ}=-RTlnK_{eq}$$, the natural log must be of a number greater than $$1$$ so that the right side and the left side of the equation can both be negative. By the equation $$\Delta G^{\circ}=-nFE^{\circ}_{cell}$$, the left side is negative, so the right side must be negative, which is only true if $$E^o>0$$. Thus, the reaction is exothermic because $$\Delta H<0$$, causes matter to be more dispersed and/or of higher energy since $$\Delta S>0$$, is thermodynamically favorable $$(\Delta G^{\circ}<0)$$ at all temperatures, and is product favored because $$K_{eq}>1$$. Since $$E^o>0$$ this would be an electrochemical cell that generates a voltage, and is therefore a galvanic/voltaic cell.