Answers
Ans
The balanced reaction
HIO3 + NaOH → NaIO3 + H2O
HIO3 = weak acid
NaOH = strong base
NaIO3 = salt (Conjugate base)
H2O = neutral
Part d
Volume of HIO3 = 70 mL
Molarity of HIO3 = 0.105 M
Molarity of NaOH = 0.350 M
The dissociation reaction
HIO3 = H+ + IO3-
I 0.105
C -x +x +x
E 0.105-x x x
Equilibrium constant expression of the reaction
Ka = [H+] [IO3-] / [HIO3]
0.17 = x²/(0.105-x)
0.01785 - 0.17x - x² = 0
x = 0.0734
pH = - log [H+] = - log (0.0734)
= 1.13
Part e
Moles of HIO3 = 0.070 L x 0.105 mol/L
= 0.00735 mol
Moles of NaOH = 0.015 L x 0.350 mol/L
= 0.00525 mol
Moles of HIO3 unreacted = 0.00735 - 0.00525
= 0.0021 mol
Volume = 70 + 15 = 85 mL = 0.085 L
Concentration of HIO3 = 0.0021/0.085 = 0.0247 M
The dissociation reaction
HIO3 = H+ + IO3-
I 0.0247
C -x +x +x
E 0.0247-x x x
Equilibrium constant expression of the reaction
Ka = [H+] [IO3-] / [HIO3]
0.17 = x²/(0.0247-x)
0.004199 - 0.17x - x² = 0
x = 0.0219
pH = - log [H+] = - log (0.0219)
= 1.66
Part f
pH at equivalence point
Moles of HIO3 = 0.070 L x 0.105 mol/L
= 0.00735 mol
Moles of NaOH = 0.00735 mol
Volume of NaOH = 0.00735/0.350
= 0.021 L = 21 mL
Total volume = 70 + 21 = 91 mL
Concentration of NaIO3 = 0.00735/0.091 = 0.0807 M
HIO3 + NaOH → NaIO3 + H2O
NaIO3 + H2O = NaHIO3 + OH-
I 0.0807
C -x +x +x
E 0.0807-x x x
Equilibrium constant expression of the reaction
Kb = [NaHIO3] [OH-] / [NaIO3]
(10^-14/0.17) = x²/0.0807-x
4.75*10^-15 - 5.88*10^-14*x - x² = 0
x = 6.89*10^-8
pOH = - log (6.89*10^-8) = 7.16
pH = 14 - 7.16 = 6.84
Part g
Moles of HIO3 = 0.070 L x 0.105 mol/L
= 0.00735 mol
Moles of NaOH = 0.035 L x 0.350 mol/L
= 0.01225 mol
Moles of NaOH in excess = 0.01225 - 0.00735
= 0.0049 M
pOH = - log [OH-] = - log [0.0049]
= 2.31
pH = 14 - 2.31
= 11.69
