I was able to get part of the problem done. Id really appreciate any help unders

Question

I was able to get part of the problem done. Id really appreciate any help understanding/ getting the rest of the problem finished up!! Thanks!!

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HCIO(aq) with 0.200 M KOH(aq). The ionization constant for HCIO can be found here. Before addition of any KOH after addition of 25.0 mL of KOH after addition of 30.0 mL of KOH after addition of 50.0 mL of KOH after addition of 60.0 mL of KOH With equal concentrations of monoprotic titrant and analytic, the equivalence point would occur when the added volumes are equal (50.0 mL of KOH added). So at point (b), we are 25.0/50.0 = 50% of the way to the equivalence point. If 50% of the acid has reacted, then 50% remains, and 50% of the conjugate base has been formed so [CIO ]/[HCIO] = 50/50 = 1. Find the pH using the Henderson-Hasselbalch equation. pH = pKa + log [ cio- ]/ [HCIO ]

Explanation / Answer

pH = pKa + log [salt/acid]

pKa of HClO = 7.1

B) after addition of 25 ml of KOH

concentration of KOH = 0.2*25/75 = 0.07 M

concentration of HClO = 0.2*50/75 = 0.133 M

net concentration of HClO = 0.133-0.7 = 0.063 M

pH    = pKa+log [base/acid]

   = 7.1 + log (0.07/0.063)

   = 7.146

C) after addition of 30 ml of KOH

concentration of KOH = 0.2*30/80 = 0.075 M

concentration of HClO = 0.2*50/80 = 0.125 M

net concentration of HClO = 0.125-0.075 = 0.05 M

pH    = pKa+log [base/acid]

   = 7.1 + log (0.075/0.05)

   = 7.276
e) after 60 ml of KOH added

at this point pH of solution depends upon excess KOH added

at equivalence 50 ml KOH used up .so that remaining KOH is 10ml.

concentration of excess KOH = 10*0.2/110 = 0.018

pOH = -log 0.018   = 1.745

pH = 14-1.745 = 12.26

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