Question

what will pH at equivalence point in titration of 45.00 mL .115 M NaClO with .106 M HCl?

Answer 1

NaClO(aq) + HCl(aq) ----> HClO(aq) + NaCl(aq)

no of mol of NaClO = 45*0.115 = 5.175 mol

no of mol of HCl required = 5.175 mol

vol of HCl required =n/M = 5.175/0.106 = 48.82 ml

concentration of HClO formed = n/V = 5.175/(45+48.82) = 0.5516 M

pH of HClO = 1/2(pka-logC)

pka of HClO = 7.54

   Concentration of HClO = 0.5516 M

             = 1/2(7.54-log0.5516)

             = 3.9