Question

A buffer solution contains 0.34 mol of phenol (HC₆H₅O) and 0.87 mol of sodium phenoxide (NaC₆H₅O) in 2.60 L.
The K_a of phenol (HC₆H₅O) is K_a = 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 =  

Answer 1

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