Questions: Test Yourself: Find the pH if we add another 1.00 g of tris hydrochloride. [B]=12.43 g / L / 121.14 g / mol = 0.1026 M [BH+]=4.67 g / L / 157.59 g / mol = 0.0296 M Henderson-Hasselbalch Equation The Henderson-Hasselbalch equation is a rearranged form of the Ka equilibrium expression. Ka=[H+][A-]/[HA] Henderson-Hasselbalch equation for an acid: HA ↔ H+ + A- pH = pKa + log ([A-]/[HA]) Henderson-Hasselbalch equation for a base: Ka=Kw / Kb BH+ ↔ B + H+ pH = pKa + log ([B]/[BH+]) Note: If activities are included, pH = pKa + log ([A-] γA-]/[HA] γHA) pKa applies to this acid

Transcript text: Test Yourself: Find the pH if we add another 1.00 g of tris hydrochloride. \[ [B]=\frac{12.43 \mathrm{~g} / \mathrm{L}}{121.14 \mathrm{~g} / \mathrm{mol}}=0.1026 \mathrm{M}\left[\mathrm{BH}^{+}\right]=\frac{4.67 \mathrm{~g} / \mathrm{L}}{157.59 \mathrm{~g} / \mathrm{mol}}=0.0296 \mathrm{M} \] Henderson-Hasselbalch Equation The Henderson-Hasselbalch equation is a rearranged form of the $K_{\mathrm{a}}$ equilibrium expression. \[ K_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \] Henderson-Hasselbalch equation for an acid: \[ \begin{array}{c} \mathrm{HA} \stackrel{K_{\mathrm{a}}}{\rightleftharpoons} \mathrm{H}^{+}+\mathrm{A}^{-} \\ \mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \end{array} \] Henderson-Hasselbalch equation for a base: \[ \begin{array}{c} K_{\mathrm{a}}=K_{\mathrm{w}} / K_{\mathrm{b}} \\ \mathrm{BH}^{+} \leftrightharpoons \mathrm{B}+\mathrm{H}^{+} \end{array} \] \[ \mathbf{p H}=\mathbf{p} \boldsymbol{K}_{\mathrm{a}}+\log \frac{[\mathrm{B}]}{\left[\mathbf{B H}^{+}\right]} \] Note: If activities are included, $\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right] \gamma_{\mathrm{A}^{-}}}{[\mathrm{HA}] \gamma_{\mathrm{HA}}}$ $\mathrm{p} K_{\mathrm{a}}$ applies to this acid