Question

How will the equivalence point volume change if you titrate the two solutions (solution 1: a 10mL vinegar solution that has a concentration of 5%(w/v%))(solution 2: a 10-mL vinegar solution that has a concentration of 5% (w/v%) together with 30mL of water). What is the pH of the equivalence point of the two solutions if you titrate with 0.3M NaOH?

Answer 1

solution 1: pH at equivalence point = 7+1/2(pka+logC)

                               pka of aceticacid = 4.74

No of mol of aceticacid = (10*5/100)/60 = 0.00833 mol

No of NaOH required = 0.00833 mol

Volume of NaOH required = 0.00833/ 0.3 = 0.028 L = 28 ml

total volume of mixe r = 28+10 = 38 ml

concentration of salt = 0.00833/0.038= 0.22 M

pH = 7+1/2(4.74+log0.22) = 9.04

solution ; 2

pH at equivalence point = 7+1/2(pka+logC)

                               pka of aceticacid = 4.74

No of mol of aceticacid = (10*5/100)/60 = 0.00833 mol

No of NaOH required = 0.00833 mol

Volume of NaOH required = 0.00833/ 0.3 = 0.028 L = 28 ml

total volume of mixe r = 28+10+30 = 68 ml

concentration of salt = 0.00833/0.068= 0.1225 M

pH = 7+1/2(4.74+log0.1225) = 8.914