Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.240 M HClO(aq) with 0.240 M KOH(aq).

The ionization constant for HClO can be found here.

https://sites.google.com/site/chempendix/ionization

(a) before addition of any KOH

(b) after addition of 25.0 mL of KOH

(c) after addition of 30.0 mL of KOH

(d) after addition of 50.0 mL of KOH

(e) after addition of 60.0 mL of KOH

Answer 1

pka of HClO = -LOG(4*10^(-8)) = 7.4

(a) before addition of any KOH

Ph = 1/2(PKA-LOGc)

   = 1/2(7.4-log0.24) = 4

(b) after addition of 25.0 mL of KOH

No of mol of HClO = 50/1000*0.24 = 0.012 mol

No of mol of KOH added = 25/1000*0.24 = 0.006 mol

at half equivalence point

pH = pka = 7.4

(c) after addition of 30.0 mL of KOH

No of mol of HClO = 50/1000*0.24 = 0.012 mol

No of mol of KOH added = 30/1000*0.24 = 0.0072 mol

pH = pka+log(salt/acid)

   = 7.4+log(0.0072/(0.012-0.0072))

   = 7.58

(d) after addition of 50.0 mL of KOH

No of mol of HClO = 50/1000*0.24 = 0.012 mol

No of mol of KOH added = 50/1000*0.24 = 0.012 mol

concentration of salt = 0.012/0.1 = 0.12 M

at equivalence point

pH = 7+1/2(pka+logC)

   = 7+1/2(7.4+log0.12)

   = 10.23

(e) after addition of 60.0 mL of KOH

No of mol of HClO = 50/1000*0.24 = 0.012 mol

No of mol of KOH added = 60/1000*0.24 = 0.0144 mol

excess KOH = 0.0144-0.012 = 0.0024 mol

concentration of KOH = 0.0024/0.11 = 0.0218 M

pOH = -LOG(oh-) = -log(0.0218)

     = 1.66

pH = 14-1.66 = 12.34