Question

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3. a. (35) Consider the titration of 70.00 mL 0.105 MHIO, with 0.350 M NaOH. Write the reaction which occurs during the titration Reaction Under each species in the equation, claually as a strong acid (5A). trong bato (58). weak acid (WA) or a weck base (W) or neutral (N). b. Label the diagram of a titration experiment. What solution is in the buret? What is in the beaker? PH c. Draw a titration curve that will illustrate the pH changes as sodium hydroxide is added.

Answer 1

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