Question

Can you please solve this?

A 1.0000 g sample that contains both sodium carbonate and sodium bicarbonate is dissolved in water and titrated with 0.1000 M hydrochloric acid. The burette reading at a phenolphthalein endpoint (pH 8.0-9.6) is 14.75 mL; the titration is continued to a methyl orange endpoint (pH 3.1 – 3.4), which occurs at 38.85 mL. For carbonic acid (H,CO), pK, = 6.352 and pk, = 10.329. (a) Write the reaction equations and the expressions for K and K for NaHCO, (b) Sketch the general appearance of the curve (pH verses volume of titrant) for this titration. Explain (in words) what chemistry governs the pH in each distinct region of the curve. Calculate how much sodium carbonate and sodium bicarbonate is in the sample. (c)

Answer 1

(a)

NaHCO₃ (aq) -------> Na⁺ (aq) + HCO₃^-(aq)

Reaction and expression for Ka :

   HCO₃^-(aq) + H₂O (l)$\rightleftharpoons$   H₃O⁺ (aq) + CO₃^2-(aq)

since water is in excess :

Ka = (H₃O⁺)( CO₃^2-)/( HCO₃^-)

Reaction and expression for Kb :

   HCO₃^-(aq) + H₂O (l)$\rightleftharpoons$ OH^-(aq) + H₂CO₃(aq)

since water is in excess :

Kb = (OH^-)( H₂CO₃)/( HCO₃^-)

(b) pH will be governed by various factors at various stage of titration :

1. pH at start of titration : (solution will be basic : pH : ~11)

principal equilibrium :

CO₃^2-(aq) + H₂O (l)$\rightleftharpoons$ OH^-(aq) + HCO₃^-(aq) ;   Kb = (OH^-)( HCO₃^-)/( CO₃^2-)

pH will be controlled by Kb and ( CO₃^2-) .

2.before first equivalence point :

principal equilibrium :

CO₃^2-(aq) + H₃O⁺ (aq)$\rightleftharpoons$ HCO₃^-(aq) + H₂O (l)

It will behave as buffer of ( CO₃^2-) and( HCO₃^-) : equation for pH of buffer will be used (pKa2).

3.pH at 1st equivalence point :(phenolphthalein end point )

pH = 1/2 (pKa1 + pKa2) = 8.3

4. between 1st and 2nd equivalence point :

principal equilibrium :

HCO₃^-(aq) + H₃O⁺ (aq)$\rightleftharpoons$ H₂CO₃(aq) + H₂O (l)

equation for pH of buffer will be used (pKa1).

5. 2nd equivalence point : excess of HCl will determine pH.

(c) reaction during titration :

CO₃^2-(aq) + H₃O⁺ (aq)$\rightleftharpoons$ HCO₃^-(aq) + H₂O (l) (phenolphthalein end point : V₁)

HCO₃^-(aq) + H₃O⁺ (aq)$\rightleftharpoons$ H₂CO₃(aq) + H₂O (l) (methyl orange end point : V)

1.First point is solely due to Na₂CO₃ , so volume of HCl used for reacting with it : 2*V₁

= 2*14.75 ml = 29.5 ml

moles of HCl reacted = 0.1000 mol/L * 29.5 ml / 1000 ml/L = 0.00295 moles

1 mol Na₂CO₃ = 2 mol of HCl

moles of Na₂CO₃ present = 0.001475 mole

mass of Na₂CO₃ present in sample = 0.001475 mole * 106 g/mol = 0.1564 g

2.

1. Volume of HCl reacted with NaHCO₃ : V -2V₁ = 9.35 ml

moles of HCl reacted = 0.1000 mol/L * 9.35 ml / 1000 ml/L = 0.000935 moles

1 mol NaHCO₃ = 1 mol of HCl

moles of Na₂CO₃ present = 0.000935 moles

mass of NaHCO₃ present in sample = 0.000935 moles * 84 g/mol = 0.0786 g