A buffer solution contains 0.34 mol of phenol
(HC6H5O) and 0.87 mol of sodium phenoxide
(NaC6H5O) in 2.60 L.
The Ka of phenol (HC6H5O) is
Ka = 1.3e-10.
(a) What is the pH of this buffer?
pH =
(b) What is the pH of the buffer after the addition of 0.12 mol of
NaOH? (assume no volume change)
pH =
(c) What is the pH of the original buffer after the addition of
0.10 mol of HI? (assume no volume change)
pH =
a)
Ka = 1.3*10^-10
pKa = - log (Ka)
= - log(1.3*10^-10)
= 9.886
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.886+ log {0.87/0.34}
= 10.29
Answer: 10.29
b)
mol of NaOH added = 0.12 mol
HC6H5O will react with OH- to form C6H5O-
Before Reaction:
mol of C6H5O- = 0.87 mol
mol of HC6H5O = 0.34 mol
after reaction,
mol of C6H5O- = mol present initially + mol added
mol of C6H5O- = (0.87 + 0.12) mol
mol of C6H5O- = 0.99 mol
mol of HC6H5O = mol present initially - mol added
mol of HC6H5O = (0.34 - 0.12) mol
mol of HC6H5O = 0.22 mol
Ka = 1.3*10^-10
pKa = - log (Ka)
= - log(1.3*10^-10)
= 9.886
since volume is both in numerator and denominator, we can use mol
instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.886+ log {0.99/0.22}
= 10.54
Answer: 10.54
c)
mol of HI added = 0.1 mol
C6H5O- will react with H+ to form HC6H5O
Before Reaction:
mol of C6H5O- = 0.87 mol
mol of HC6H5O = 0.34 mol
after reaction,
mol of C6H5O- = mol present initially - mol added
mol of C6H5O- = (0.87 - 0.1) mol
mol of C6H5O- = 0.77 mol
mol of HC6H5O = mol present initially + mol added
mol of HC6H5O = (0.34 + 0.1) mol
mol of HC6H5O = 0.44 mol
Ka = 1.3*10^-10
pKa = - log (Ka)
= - log(1.3*10^-10)
= 9.886
since volume is both in numerator and denominator, we can use mol
instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 9.886+ log {0.77/0.44}
= 10.13
Answer: 10.13
A buffer solution contains 0.34 mol of phenol (HC6H5O) and 0.87 mol of sodium phenoxide (NaC6H5O)...
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