DETERMINATION: Soda Ash Determination You will prepare roughly 0.1 M HCI. It wil
ID: 944786 • Letter: D
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DETERMINATION: Soda Ash Determination You will prepare roughly 0.1 M HCI. It will be standardized using primary standard sodium carbonate. Then the sodium carbonate unknown is titrated with the standardized HCl. The equations below apply to this titration. [ H30+ + CO32-_ HCO3-+ H2O Titration by HCL 1 H3O+ + HCO3- H2CO3 Boiling: H2CO3 H2O + CO2(g) The unknown carbonate is dissolved in water and titrated to a Bromcresol Green endpoint, pH around 4. This endpoint is rather vague because of the presence of the buffering due to protonated carbonate species + CO2 just before the equivalence point. Boiling removes essentially all the H2CO3 and CO2. This leaves just a small amount of HCO3 which is readily titrated to a sharp endpoint with this indicator. Procedure Make up 1 L of 0.1 M HCI by dissolving 8-9 mL conc. HCI in a 1 L bottle, fill with deionized water and shake well. HCI Standardization: Obtain 2 g primary standard Na CO3 and dry it for 2 hours at 110°C. Weigh out 0.2 g samples accurately into titration flasks. Dissolve in-75 mL. boiled water and add 4 drops of Bromcresol Green. Titrate to a greenish color with your standardized 0.1 MHCI. Boil the solution gently, without splashing. Cover the flask and cool. Then continue the titration to the endpoint. If additional acid is not required after boiling then discard the sample Sodium Carbonate Determination. Dry the sample for two hours at 110 °C. Weigh out an-0.4 g trial sample accurately into a titration flask. Dissolve in -75 mL. boiled water and titrate as in the standardization reaction Successive sample sizes should be chosen to require around 35 mL of acid. Report the % Na2CO3 in the sample, and its r.s.d.Explanation / Answer
moles of HCl used for titration = 0.1 M x 35 ml = 3.5 mmol
moles of CO3^2- present = 3.5/2 = 1.75 mmol
grams of Na2CO3 present in the unknown = 1.75 mmol x 106 g/mol = 0.1855 g
Original weight of unknown sample = 0.4 g
wt% of Na2CO3 in sample = 0.1855 x 100/0.4 = 46.375%
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