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 ? Ka HOBr = 2.5 x 10-9
What is the pH of a 1.14M solution of carbonic acid? Ka1 = 4.3 x 10-7 Ka2 = 5.6 x 10-11
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
What is the pH after 0.28 moles of HCl is added to a buffer containing 1.00...
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