pH = -log10[H+] Weak acid: pH = IONIC PRODUCT OF WATER: Ka = [H+][OH-]/[H2O] as conc of water is near constant, it’s not included so… Kw =[H+][OH-] Kw = 1x10-14 (roughly) For pure water: [H+] = [OH-] pKa HA H+ + A- pKa = -log10(Ka) Ka = [H+][A-]/[HA] Strong (hydrochloric) nearly completely dissociate. Large Ka and therefore a small pKa Weak acids (acetic) are mainly undissociated. Smaller Ka and therefore a larger pKa More dissociated, equilibrium shifts to the RHS = larger Ka = smaller pKa Describe what buffers are, give examples, and perform simple calculations relating to use Buffer solution: a solution that resists changes in pH on the addition of small amounts of acid/base Weak acid and bases are able to buffer the pH of a solution over a range of 1 either side of pKa. HENDERSON-HASSELBALCH EQUATION HA H+ + A- Equation can be used to estimate the pH of a solution cantaining an acid and its conjugate base. When [A-] = [HA], pH = pKa Therefore half the acid is ionised when pH = pKa
pH = -log10[H+] Weak acid: pH = IONIC PRODUCT OF WATER: Ka = [H+][OH-]/[H2O] as conc of water is near constant, it’s not included so… Kw =[H+][OH-] Kw = 1x10-14 (roughly) For pure water: [H+] = [OH-] pKa HA H+ + A- pKa = -log10(Ka) Ka = [H+][A-]/[HA] Strong (hydrochloric) nearly completely dissociate. Large Ka and therefore a small pKa Weak acids (acetic) are mainly undissociated. Smaller Ka and therefore a larger pKa More dissociated, equilibrium shifts to the RHS = larger Ka = smaller pKa Describe what buffers are, give examples, and perform simple calculations relating to use Buffer solution: a solution that resists changes in pH on the addition of small amounts of acid/base Weak acid and bases are able to buffer the pH of a solution over a range of 1 either side of pKa. HENDERSON-HASSELBALCH EQUATION HA H+ + A- Equation can be used to estimate the pH of a solution cantaining an acid and its conjugate base. When [A-] = [HA], pH = pKa Therefore half the acid is ionised when pH = pKa
CBB-Notes.docx