1. Acid-Base Titrations

2. Titrations TITRATION is the process of determining the concentration of a solution by reacting it with a solution of a known concentration. To determine the unknown concentration, we must measure the volume of acid or base carefully using a burette (see figure). By recording the initial volume and final volume, the volume of solution added can be easily measured.

3. Tools for Titrations Pipette Burette Indicator Acid Base Erlenmeyer flask

4. Titration Terms Aliquot – A portion of a solution. Equivalence Point – The point where you have added stoichiometric amounts of the acid or base, ie. neutralization occurs. End Point – The point where the indicator changes colour. – In a good titration the end point will be very close to the equivalence point.

5. Reading a Burette A burette scale is inverted. Individual readings only make sense when compared to a second value. V initial = 12.45 mL V final = 15.75 mL V added = 3.30 mL

6. Example 1 – Titration of a Strong Acid with a Strong Base A 10.00 mL solution of hydrochloric acid was titrated with 0.225 mol/L sodium hydroxide. The following data was collected: Burette Readings: V initial = 12.45 mL V final = 15.75 mL Volume Added: V added = V final – V initial = 3.30 mL

7. Step 1: Balanced Chemical Equation (BCE) HCl (aq) +NaOH (aq)NaCl (aq) + H 2 O Givens & unknown data: c acid = ? c base = 0.225 mol/L V acid = 10.00 mL = 0.01000 L V base = 3.30 mL = 0.00330 L

8. Step2: Moles Base: Since mole ratio is 1:1 then, n acid = n base = 7.425 x 10 -4 mol Step 3: Mole Ratios n base = c b V b = (0.00330)(0.225) = 7.425 x 10 -4 mol

9. Step 4: Concentration of Acid c acid = n a /V a = (7.425 x 10 -4 mol)/(0.01000L) = 0.0742 mol/L Therefore the concentration of acid was 0.0742 mol/L.

10. Example 2 – Titration of a Strong Base with a Diprotic Acid What volume of 0.159 mol/L KOH solution would be required to neutralize 27.48 mL of 0.104 mol/L H 2 SO 4 ? BCE: H 2 SO 4 (aq) + 2 KOH (aq) K 2 SO 4 (aq) + 2H 2 O C acid = 0.104 mol/L c base = 0.159 mol/L V acid = 27.48 mL = 0.02748 L V base = ? Moles Acid: n acid = cV = (0.104)(0.02748) = 0.002858 mol Mole Ratio n base => 2 = ____n base ___ n acid 1 0.002858 mol n base = 2 x n acid = (2) (0.002858) = 0.005716 mol Volume of Base Needed: V = n/c = 0.005716 / 0.159 = 0.035948 L V = 35.9 mL

11. Example 3 – Cleaning up an Acid Spill What mass of sodium hydrogen carbonate is needed to clean up a 250 mL spill of concentrated hydrofluoric acid (27.6 mol/L)? BCE: NaHCO 3 (s) + HF (aq) NaF (aq) + H 2 O+ CO 2 (g) m = ?V acid = 250 mL M = 84.01 g/molc acid = 27.6 mol/L Moles Acid:n acid = cV = (27.6 mol/L)(0.250 L) = 6.90 mol Mole Ratio: n acid = n base = 6.90 mol Moles to Mass: m = nM = (6.90 mol)(84.01 g/mol) = 580 g Therefore 580 g of the sodium hydrogen carbonate is required for the clean-up.

12. CLASSWORK/HOMEWORK Complete Acid-Base Reactions and Titration Calculations Worksheet