Consider a mixture that may contain NaOH (Mw = 40.0), Na_2CO_3 (Mw = 106) and Na

Question

Consider a mixture that may contain NaOH (Mw = 40.0), Na_2CO_3 (Mw = 106) and NaHCO_3 (Mw = 84.0) along with inert material. A 2.000 g sample containing this mixture was dissolved in water and diluted to 100.0 mL in a volumetric flask. Titration of a 50.0 mL portion to a phenolphthalein end point requires 29.64 mL of 0.1202 M HCl. Another 50.0 mL portion from the volumetric flask requires 36.42 mL of the 0.1202 M HCl when titrated to a bromocresol green end point a) Which species are present? b) Calculate the percent of each species present in the original sample.

Explanation / Answer

In the qualitative and quantitative determinations solution containing Na2CO3, NaHCO3 and NaOH alone or in a mixture, no more than two of these constituents can be present in a solution together since the reaction between some of them eliminates the third one.

Analysis of these mixtures require the use of two indicators such as phenolphthalein (for detecting the first end point) and an acid range indicator such as bromocresol green (for the second end point).

There is a relationship avavilable in the analysis of carbonate and carbonate mixtures during titration

(a) since Vph (29.64 mL) < Vbcg (36.42 mL), so, the solution has the mixture of Na2CO3 and NaHCO3 .

(b) mass of components in the unknown solutions can be calculated as follows,

(i) mg of Na2CO3 = Vph x MHCl x formula weight of Na2CO3

= 29.64 x 0.1202 x 106 = 377.64 mg = 0.377 g

(ii) mg of NaHCO3 = (Vbcg – Vph) x MHCl x formula weight of NaHCO3

= (36.42-29.64) x 0.1202 x 84

= 9.78 x 0.1202 x 84 = 98.64 mg = 0.098 g

% of NaHCO3 = (0.377 g /2 g) x 100 = 18.85 %

% of Na2CO3 = (0.098 g/2 g) x 100 = 4.9 %

composition in the solution relationship between Vconsumed after phenolphthalein end point and Vconsumed after bromocresol green end point NaOH Vphth>0, Vbcg=0 Na2CO3 Vphth = Vbcg NaHCO3 Vph=0 Vbcg>0 NaOH and Na2CO3 Vph > Vbcg Na2CO3 and NaHCO3 Vph< Vbcg

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