This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's a breakdown of the concepts and derivations shown in your notes: Step 1: Definition of pKa and Henderson-Hasselbalch Equation for Acids The acid dissociation constant, Ka, describes the extent of dissociation of a weak acid. pKa is defined as the negative logarithm of Ka. pKa = - Ka For a weak acid HA dissociating into H^+ and its conjugate base A^-: HA H^+ + A^- The equilibrium constant Ka is given by: Ka = ([H^+][A^-])/([HA]) Taking the negative logarithm of both sides: Ka = - (([H^+][A^-])/([HA])) pKa = - [H^+] - (([A^-])/([HA])) Since pH = - [H^+], we can substitute this into the equation: pKa = pH - (([A^-])/([HA])) Rearranging to solve for pH, which is the Henderson-Hasselbalch equation for acids: pH = pKa + (([A^-])/([HA])) In a buffer solution, [A^-] is often represented as the concentration of the salt (conjugate base) and [HA] as the concentration of the acid. pH = pKa + ([Salt])/([Acid]) Step 2: Henderson-Hasselbalch Equation for Bases For a weak base BOH dissociating into B^+ and OH^-: BOH B^+ + OH^- The base dissociation constant Kb is given by: Kb = ([B^+][OH^-])/([BOH]) Similar to pKa, pKb is defined as: pKb = - Kb Taking the negative logarithm of the Kb expression: Kb = - (([B^+][OH^-])/([BOH])) pKb = - [OH^-] - (([B^+])/([BOH])) Since pOH = - [OH^-], we substitute this into the equation: pKb = pOH - (([B^+])/([BOH])) Rearranging to solve for pOH, which is the Henderson-Hasselbalch equation for bases: pOH = pKb + (([B^+])/([BOH])) In a buffer solution, [B^+] is often represented as the concentration of the salt (conjugate acid) and [BOH] as the concentration of the base. pOH = pKb + ([Salt])/([Base]) Step 3: Relationship between pH, pOH, and Kw The ion product of water, Kw, is a constant that describes the autoionization of water. At 25^ C, its value is 1.0 × 10^-14. Kw = [H^+][OH^-] = 1.0 × 10^-14 Taking the negative logarithm of Kw: Kw = - ([H^+][OH^-]) pKw = - [H^+] - [OH^-] Since pH = - [H^+] and pOH = - [OH^-]: pKw = pH + pOH At 25^ C, pKw = - (1.0 × 10^-14) = 14. Therefore, the relationship between pH and pOH is: pH + pOH = 14 Got more? Send 'em 📸