Question

Acid base titration determination of Na2CO3 in Soda Ash

CO3^2- +H⁺ = HCO3^-

HCO3^- + H⁺ = H2CO3

Buffer by mixing acetic acid and sodium acetate.

The comparison indicator solution should match the analytical solution at the second equivalence point in three respects: pH, Ionic strength, color and intensity

A) Calculate the concentration of sodium acetate necessary to achieve the ionic strength in pre lab 4. ( Pre lab 4 calulation, the ionic strength that I have calculated is to be 0.1)

B) Calculate the concentration of acetic acid given the values calculated for A.

C) Calculate the mass of unknown to weight out in order to consume 40 ml of 0.1 M HCL assuming the unknow is 35% by mass Na2CO3.

For preview calculation where you calculate the mass of Na2CO3 needed to consume 40 mL of 0.10 M HCL in a titration of Na2CO3 with HCL come out to be 0.212 gram needed to consume 40 mL of 0.10 M of HCL.

Answer 1

A clarification first: you have written Na₂CO₃ throughout but mentioned sodium acetate. Which one is correct? Do these relate to the same question or different questions? I will go with sodium acetate for the first two and sodium carbonate, Na₂CO₃ for the last one.

A) The ionic strength of the solution is μ = 0.1. We know that ionic strength

μ = 1/2Σc_iz_i² where c_i is the concentration of the i-th ion in the solution and z_i = charge of the i-th ion. For sodium carbonate, we have,

CH₃COO^-Na⁺ ------> Na⁺ + CH₃COO^-

Let c be the molar concentration of sodium acetate. Therefore, concentration of Na⁺ = c and concentration of CH₃COO^- = c (assuming complete ionization).

Therefore,

0.1 = ½*[(c)*(1)² + (c)*(-1)²] = ½*[c + c] = ½*2c = c

====> c = 0.1

Therefore, molar concentration of sodium acetate = 0.1 mol/L (ans).

B) Are you supposed to consider complete reaction of sodium acetate with an acid to form acetic acid or the pH of the buffer solution is given? Please specify.

C) Write down the reaction between Na₂CO₃ and HCl.

Na₂CO₃ + 2 HCl ------> 2 NaCl + CO₂ + H₂O

The molar ratio of reaction between Na₂CO₃ and HCl is 1:2.

Moles HCl added = (volume of HCl added)*(concentration of HCl in mol/L) = (40 mL)*(1 L/1000 mL)*(0.1 mol/L) = 0.004 mole.

Moles Na₂CO₃ reacted = (0.004 mole HCl)*(1 mole Na₂CO₃/2 mole HCl) = 0.002

Molar mass of Na₂CO₃ = 105.99 g/mol.

Therefore, mass of Na₂CO₃ taken = (0.002 mole)*(105.99 g/1 mol) = 0.21198 g ≈ 0.212 g.

This would have been the weight of Na₂CO₃ required if the unknown was 100% Na₂CO₃. However, the unknown contains 35% Na₂CO₃ as per the question.

Let the mass of unknown taken be x g.

Therefore, (35/100)*x = 0.212

====> x = 0.212*100/35 = 0.6057 ≈ 0.606

The mass of unknown taken = 0.606 g (ans).