Question

Consider 500 mL of a buffer comprised of 0.12 M HOCl and 0.080 M KOCl.

a) Do you predict the pH of the buffer to be above or below the pKa? (Please explain)

b)Given that HOCl has a Ka= 3.5x10^8- , Calculate the pH of this buffer? Does your calculated value match your prediction?

c)Let's add 2.5 mL of 10 M KOH to this buffer. Do you predict the pH of the buffer to increase, decrease or stay the same whe you add KOH? (Please explain)

d)Calculate the pH of the buffer after the addition of hte strong base. (Please show all work)

Answer 1

since pH= pKa+ log [Conjugate base/acid]

given Ka= 3.5*10^-8, pKa= 7.45

if pH = 7.45+ log (0.08/0.12)= 7.27

hence pH< PKa

2. The addition of KOH leads to HOCl reacting with KOH to form more of KOCl according to

HOCl+ KOH----->KOCl+ H2O, 1 mole of HOCl requires 1 mole of KOH for completely react

moles of KOH added= molarity* volume in L= 10*2.5/1000 =0.025 moles

moles of HOCL in 0.5L of 0.12M= 0.12*0.5= 0.06 moles,

since KOH is limiting , all the KOH reacts and gives rise to additionally 0.025 moles of KOCl.

so total moles of KOCl= 0.08*0.5+0.025=0.065, moles of HOCl remaining = 0.06-0.025=0.035

volume of solution now = 500+2.5= 502.5 ml =0.5025L

concentrations (M) : HOCl= 0.035/0.5025=0.07 and KOCl= 0.065/0.5025

new pH= 7.45+ log (0.065/0.07)=7.42 , there is an increase in pH from 7.27 to 7.42