Question

What is the pH after 0.28 moles of HCl is added to a buffer containing 1.00 moles of HOBr and 1.00 moles NaOBr ? K_{a HOBr} = 2.5 x 10^-9

What is the pH of a 1.14M solution of carbonic acid? K_a1 = 4.3 x 10^-7  K_a2 = 5.6 x 10^-11

Answer 1

Ka = 2.5 x 10^-9

pKa = -logKa = -log (2.5 x 10^-9) = 8.60

1 mol HOBr + 1 mol NaOBr is a buffer . to this buffet additionally 0.28 mole HCl is added

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

on addition of ’ C’ moles of acid to acidic buffer salt moles decreases and acid moles increases

so

pH = pKa + log [Salt –C/acid + C]

pH = 8.60 + log [1-0.28 / 1+ 0.28]

pH = 8.35

2 )

H2CO3 -------------------------> H+   + HCO3-

1.14                                       0            0

1.14-x                                   x             x

Ka1 = x^2 / 1.14-x

4.3 x 10^-7 = x^2 / 1.14-x

x^2 + 4.3 x 10^-7 x - 4.9 x 10^-7 = 0

x = 7 x 10^-4

x = [H+] = 7 x 10^-4 M

pH = -log[H+]

pH = -log ( 7 x 10^-4)

pH = 3.15