Question

Extra credit Homework on Acid-Base Titration You can use the technique of titration to determine the concentration of a mixture of sodium carbonate and sodium bicarbonate. Titration is performed using a solution with a known concentration of hydrochloric acid (HCI). During the titration, HCl reduces gradually the alkalinity of the solution until the pH is acid. The reaction occurs in two stages in the first part you titrate the total amount of carbonates present in the mixture turning them into bicarbonate. In the second stage of the titration you titrate the total amount of bicarbonate which adds the amount that comes from carbonate and the initial amount present the sample. In base to this explanation answer: 1. Write the equations for the two titrations performed in stage 1 and 2 2. Mention the two indicators that we used in this titration. What is the range of pH that each indicator has for their color change? 3. A sample of 2.0 g of Sodium carbonate, sodium bicarbonate and a neutral substance is titrated with HCI 0.5 M. Fill the table according to the volumes used during the titration Initial Volume of HCl (ml) 0 Final Volume of HCI (ML) 14.5 15.5 Total volume HCI Stage 1 Total Volume HCI Stage 2 (mL) Moles of HC Stage 1 Moles of Na.co Mass of Na,CO, % of Na.Co, Present Moles of NaHCO, Titrated (Stage 2) Moles of NaHCO, Init. Present Mass of NaHCO, Init. Present % of NaHCO, Init. Present

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Answer 1

1. Stap 1: C05 + H, HOO, Stage: 100,+H* > 409,=> 14.07.60 7 2 Shang 1. Phenolattalion (pH= 8-q-8) Stage 2: Mange Orange (blt= 3:0~416) Total Volume Hel Stage 1 : 415 mL l 15.5mL

! ! Modes of Hel. Stage 1 ): 0-0073 0.0078 Moles of the Mol Naco 0.0023 0.0078 (Stage 2). Mass of Ne, CO : (0*072*105*78-0774 (6078-105-99)-0837 (39:274-5) | (378-15:5) * 0-2005 LOCOBS 7.8 No.com : (6324,00) - 38.7% pored)=41-357 =00119 ) = 0 olla Moles of Nalco, 00119 0.011 titrated (Stage 2) (0-0111-00078) - och Natico, init prosent :Q:0119-8.0073) Mass of Nahe init. prosent : 6-00163299 % of Naltco, init. present present, (0-00963489) = 0.3869, (0.0033x84)=0-2779. 19.3% 13.85 %. O 0-0046