Recipes for Acid-base Titration

Recipes for acid-base titration General Chemistry Contents Introduction Here is everything you need to do simple calculations of pH in acid-base titrations. For each kind of titration, there are four circumstances that need to be handled differently, according to whether you are at • the start of the titration, • before the equivalence point, • at the equivalence point, or • after the equivalence point. It is crucial that you first calculate the reaction of the acid and base as a limiting reagent problem, using moles of acid and base, to determine which of these four circumstances you have. Then, you need to set up and solve the equilibrium equations for that circumstance, using concentrations. As you study these notes, try to master the method rather than to memorize formulas. It is easy to write down the formulas you need once you understand the method. 1 Titration of strong acid with strong base The first thing to do — always — is to calculate the reaction of the acid and base, H3 O+ (aq) + OH− (aq) -→ H2 O(l) + H2 O(l), as a limiting reagent problem (ignore the anion of the acid and the cation of the base). Do this by computing the number na of moles of acid and the number nb of moles of base. Then, • If nb = 0, we have pure strong acid. The pH is pH = − log ca . • If nb 6= 0 but nb < na , we have a unreacted acid. The reaction of the base leaves na − nb moles of unreacted acid, so that ca = na − nb Va + Vb and hence pH = − log ca . • If nb = na , we have a pure water. The reaction of the base produces na −nb = 0 moles of unreacted acid and nb − na = 0 moles of unreacted base, so that pH = 7. • If nb > na , we have a excess base. The reaction of the base consumes all of the acid and leaves nb − na moles of unreacted base, so that cb = nb − na Va + Vb and hence pH = 14 + log cb . 2 2 Titration of weak acid with strong base The first thing to do — always — is to calculate the reaction of the acid and base, HA(aq) + OH− (aq) -→ H2 O(l) + A− (aq), as a limiting reagent problem (ignore the cation of the strong base). Do this by computing the number na of moles of acid and the number nb of moles of base. Then, • If nb = 0, we have pure weak acid. The equilibrium relation is ← H3 O+ (aq) + A− (aq). HA(aq) + H2 O(l) → Ka = p x2 x2 =⇒ x ≈ Ka ca ≈ ca − x ca and so 1 log (Ka ca ) . 2 • If nb 6= 0 but nb < na , we have a buffer. The reaction of the strong base produces nb moles of the conjugate base, A− , of the weak acid and leaves na − nb moles of unreacted weak acid, so that pH = − log x = − ca = na − nb Va + Vb and ccb = nb . Va + Vb The equilibrium relation is ← H3 O+ (aq) + A− (aq). HA(aq) + H2 O(l) → Ka = x(ccb + x) Ka ca xccb =⇒ x ≈ ≈ ca − x ca ccb and so  pH = − log x = − log Ka ca ccb  . • If nb = na , we have hydrolysis. The reaction of the strong base produces nb = na moles of the conjugate base, A− , of the weak acid and leaves na −nb = 0 moles of unreacted weak acid, so that ccb = nb . Va + Vb The equilibrium relation is ← HA(aq) + OH− (aq). H2 O(l) + A− (aq) → s Kw ccb Kw x2 x2 Kcb = = =⇒ x ≈ ≈ Ka ccb − x ccb Ka and so pH = 14 + log x = 14 +   Kw ccb 1 log 2 Ka • If nb > na , we have a excess strong base. The reaction of the strong base produces na moles of the conjugate base, A− , of the weak acid, leaves 0 moles of unreacted acid (since na − nb < 0) and leaves nb − na moles of unreacted strong base, so that nb − na cb = . Va + Vb − The excess strong base overwhelms the OH from the hydrolysis of the conjugate base, A− , of the weak acid, so we can ignore hydrolysis and compute pH = 14 + log cb . 3 3 Titration of weak base with strong acid The first thing to do — always — is to calculate the reaction of the acid and base, H3 O+ (aq) + B(aq) -→ HB+ (aq) + H2 O(l), as a limiting reagent problem (ignore the anion of the strong acid). Do this by computing the number na of moles of acid and the number nb of moles of base. Then, • If na = 0, we have pure weak base. The equilibrium relation is ← HB+ (aq) + OH− (aq). H2 O(l) + B(aq) → Kb = p x2 x2 =⇒ x ≈ Kb cb ≈ cb − x cb and so 1 log (Kb cb ) . 2 • If na 6= 0 but na < nb , we have a buffer. The reaction of the strong acid produces na moles of the conjugate acid, HB+ , of the weak base and leaves nb − na moles of unreacted weak base, so that pH = 14 + log x = 14 + cb = nb − na Va + Vb and cca = na . Va + Vb The equilibrium relation is ← HB+ (aq) + OH− (aq). H2 O(l) + B(aq) → Kb = x(cca + x) Kb cb xcca =⇒ x ≈ ≈ cb − x cb cca and so  pH = 14 + log x = 14 + log Kb cb cca  . • If na = nb , we have hydrolysis. The reaction of the strong acid produces na = nb moles of the conjugate acid, HB+ , of the weak base and leaves nb − na = 0 moles of unreacted weak base base, so that cca = na . Va + Vb The equilibrium relation is Kca ← H3 O+ (aq) + B(aq). HB+ + H2 O(l) → s Kw cca Kw x2 x2 = = =⇒ x ≈ ≈ Kb cca − x cca Kb and so pH = − log x = −   Kw ccb 1 log 2 Kb • If na > nb , we have a excess strong acid. The reaction of the strong acid produces nb moles of the conjugate acid, HB+ , of the weak base leaves 0 moles of unreacted weak base (since nb − na < 0), and leaves na − nb moles of unreacted strong acid, so that ca = na − nb . Va + Vb The excess strong acid overwhelms the H3 O+ from the hydrolysis of the conjugate acid of the weak base, so HB+ , we can ignore hydrolysis and compute pH = − log ca . 4