Question

A buffer solution contains 0.40 mol of hypoiodous acid (HIO) and 0.20 mol of sodium hypoiodite (Nalo) in 8.30 L. The Kof hypoiodous acid (HIO) is kg = 2.3e-11. (a) What is the pH of this buffer? pH = (b) What is the pH of the buffer after the addition of 0.26 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.13 mol of HI? (assume no volume change) pH =

Answer 1

a)

Ka = 2.3*10^-11

pKa = - log (Ka)

= - log(2.3*10^-11)

= 10.64

use:

pH = pKa + log {[conjugate base]/[acid]}

= 10.64+ log {0.2/0.4}

= 10.34

Answer: 10.34

b)

mol of NaOH added = 0.26 mol

HIO will react with OH- to form IO-

Before Reaction:

mol of IO- = 0.2 mol

mol of HIO = 0.4 mol

after reaction,

mol of IO- = mol present initially + mol added

mol of IO- = (0.2 + 0.26) mol

mol of IO- = 0.46 mol

mol of HIO = mol present initially - mol added

mol of HIO = (0.4 - 0.26) mol

mol of HIO = 0.14 mol

Ka = 2.3*10^-11

pKa = - log (Ka)

= - log(2.3*10^-11)

= 10.64

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 10.64+ log {0.46/0.14}

= 11.15

Answer: 11.15

c)

mol of HI added = 0.13 mol

IO- will react with H+ to form HIO

Before Reaction:

mol of IO- = 0.2 mol

mol of HIO = 0.4 mol

after reaction,

mol of IO- = mol present initially - mol added

mol of IO- = (0.2 - 0.13) mol

mol of IO- = 0.07 mol

mol of HIO = mol present initially + mol added

mol of HIO = (0.4 + 0.13) mol

mol of HIO = 0.53 mol

Ka = 2.3*10^-11

pKa = - log (Ka)

= - log(2.3*10^-11)

= 10.64

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 10.64+ log {7*10^-2/0.53}

= 9.759

Answer: 9.76