Worksheet 23 Strong Acid/Strong Base Titrations

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1 Worksheet 2 Strong Acid/Strong Base Titrations A. Initial ph This is always determined based solely on the initial concentration of the acid or base being titrated. Every mole of acid or base will produce one mole of H O or OH. Exception: Bases formed with Group II cations will form two moles of OH for every mole of base, e.g. Ca(OH) 2. ph = log[h O ] (starting with strong acid) or poh = log[oh ] and ph = 14 poh (starting with strong base) B. ph during the titration Neutralization of 1 mole of a strong acid (H ) will produce one mole of water. H OH H 2 O Neutralization of 1 mole of a strong base (OH ) will produce 1 mole of water. OH H H 2 O The ph is determined by calculating the number of moles of unreacted acid or base remaining in the solution. This is then divided by the total volume of the solution to get the concentration of the remaining acid/base. C. Equivalence point If there are equal moles of H and OH, the only remaining species (other than the spectator ions) is water. The ph of a water solution is At the equivalence point in titrations of strong acids and bases, the solution will be neutral. D. Above the equivalence point Above the equivalence point, any added strong acid or base will determine the ph. The excess H or OH divided by the volume can be used to determine the ph of the solution.

2 1. Consider the titration of 50.0 ml of 2.0 M HNO with 1.0 M KOH. At each step of the titration... a) write a reaction to show the initial reaction upon mixing b) construct an ICE table to represent the reaction (Should you use concentration or moles?) c) determine the major species present after the reaction is complete d) calculate the ph of the solution Complete these steps for the addition of 0.0 ml, 25.0 ml, 50.0 ml, 90.0 ml, ml, ml and ml of 1.0 M KOH. 0.0 ml KOH a) HNO H 2 O NO H O b) [HNO ] [NO ] [H O ] Initial Change Equil This ICE table uses units of concentration because the volume is fixed c) The major species present at equilibrium are H 2 O, NO and H O d) ph is calculated using the concentration of [H O ] H O = 2.0 M [ ] ph = log ( 2.0) = 0. 0

3 25.0 ml KOH a) The major reaction (ignoring spectator ions) is a neutralization: H OH H 2 O 2.0 mol HNO 1mol H mol H = L = mol H 1L 1mol HNO mol OH = L = 1L 1.0 mol NaOH mol OH [H ] [OH ] [H 2 O] Initial Change Equil c) The major species present at equilibrium are H 2 O, K, NO and H d) ph is calculated using the concentration of [H ] moles [ H ] = = ml ph = log 1.0 = ( ) ml KOH a) H OH H 2 O mol OH = L = mol OH [H ] [OH ] [H 2 O] Initial Change Equil c) The major species present at equilibrium are H 2 O, K, NO and H d) ph is calculated using the concentration of [H ] moles [ H ] = = ml ph = log 0.50 = 0. ( ) 0

4 90.0 ml KOH a) H OH H 2 O mol OH = L = mol OH [H ] [OH ] [H 2 O] Initial Change Equil c) The major species present at equilibrium are H 2 O, K, NO and H d) ph is calculated using the concentration of [H ] moles [ H ] = = ml ph = log = 1. ( ) ml KOH a) H OH H 2 O mol OH = L = mol OH [H ] [OH ] [H 2 O] Initial Change Equil. 0 0 c) The major species present at equilibrium are H 2 O, K, NO d) The solution has been completely neutralized, and the ph is 7

5 110.0 ml KOH a) H OH H 2 O mol OH = L = mol OH [H ] [OH ] [H 2 O] Initial Change Equil c) The major species present at equilibrium are H 2 O, K, NO and OH d) ph is calculated using the concentration of [OH ] moles [ OH ] = = ml poh = log = 1.20 ph = 14 poh ( ) = ml KOH a) H OH H 2 O mol OH = L = mol OH [H ] [OH ] [H 2 O] Initial Change Equil c) The major species present at equilibrium are H 2 O, K, NO and OH d) ph is calculated using the concentration of [OH ] moles [ OH ] = = ml poh = log 0.25 = ( ) ph = 14 poh = 1.4

6 2. Create a titration curve by plotting the ph of the solution (yaxis) vs the volume of KOH added (xaxis) for the titration described in question #1. ph of a strong acid/strong base titration ph Volume KOH added (ml). Create a titration curve by plotting the ph of the solution (yaxis) vs the volume of KOH added (xaxis) for the titration of HNO 2 from the previous worksheet. How does the curve for a strong acid titration compare to that of a weak acid titration? ph of a weak acid/strong base titration ph Volume KOH added (ml) The strong acid/strong base curve starts at a lower ph because it begins as a strong acid solution. Beyond the equivalence point, both curves have the same ph values because the ph depends on the concentration of the excess KOH. Since the same base was used for both titrations, they have the same behavior after 100 ml has been added to either acid solution. The equivalence point of the strong acid/strong base is at ph = 7, but the equivalence point of the weak acid/strong base is NOT at ph = 7. The weak acid/strong base titration curve has a buffer zone, while the strong acid/strong base titration curve does not. (There s no buffer region in the strong acid/strong base titration because there is no conjugate weak acid/weak base pair present)

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