Question

1. What is the pH of a hypoiodous acid/sodium hypoiodite buffer in which the concentration of the weak acid component is 0.129 M and the concentration of the conjugate base is 0.102 M?Ka=2.3E-11.

2. The pH of a hypoiodous acid/sodium hypoiodite buffer in which the concentration of the weak acid component is 0.140 M and the concentration of the conjugate base is 0.070 M is 10.34. Suppose 0.128 g of solid NaOH are added to 100 mL of this buffer solution. What is the pH after the addition? Ka=2.3E-11.

Answer 1

1.

Concentration of Hypoiodous acid= 0.129M

sodium hypoiodite = 0.102M

Ka= 2.3x10^-11

-log(Ka) = -log(2.3x10^-11)

Pka = 10.64

PH= Pka + log[salt]/[acid]

PH= 10.64 + log(0.102/0.129)

PH= 10.54

2)

concentration of Hypoiodous acid = 0.140M

concentration of sodium iodite = 0.070M

volume = 100 ml

number of moles of acid = 0.140Mx0.100L=0.014 moles

number of moles of conjugate base = 0.070M x0.100L = 0.007 moles

mass of NaOH= 0.128 grams

molar mass of NaOH= 40.0 grams/mole

number of moles of NaOH= 0.128/40.0=0.0032 moles

after addition of NaOH

number of moles of acid = 0.014 - 0.0032 =0.0108moles

number of moles of conjugate base = 0.007 + 0.0032 =0.0102 moles

Ka= 2.3x10^-11

PKa = 10.64

PH= 10.64 + log(0.0102/0.0108)

PH=10.61.