CTC 11 ACID/BASE TITRATIONS Label Validation of Antacids 1. Prepare 200 ml of o.
ID: 958977 • Letter: C
Question
CTC 11 ACID/BASE TITRATIONS Label Validation of Antacids 1. Prepare 200 ml of o.l N NaOlH using dilute (6 N) NaoH. Clean a buret, rinse and fill it with the 0.1 N NaOH 3. Weigh three 0.1 g (100 mg) KHP (potassium hydrogen phtbalate . q .- 1009 I O 204.23) to the nearest 0.1 mg decimal places), Rinse the three standards into three separate 125 ml erlenmeyer flasks,add 20 to 25 ml DI and several drops of phth Titrate each standard with the NaOH to a pink endpoint. I 4. 5. 6. Calculate the normality of the NaOH. N NaOH mgkur / eq.wt.kiir, mivon Prepare 200 ml of 0.15 N HCI using dilute (6 N) HCI. 25 ml DI and several drops of fhth Titrate the HCI with the standardized NaOH to a pink endpoint. Calculate the normality of the HCL (N1 X v, 1. 2. Pipet 5 ml of O.15 N HCl into three separate 125 ml erlenmeyer flasks add 20 or BTB, 3. N2 X V:) Antacid Analysis loon logs 1. Weigh two 0.10 g (100 mg) samples of CaCO, to the nearest 0.1 mg and transfer to separate flasks with 15-20 ml DI and add phth. 2. Weigh 4 to 6 antacid tablets and calculate the average tablet weight (ATW), 12 73 3. Crush an antacid tablet in a mortar and weigh four 0.2 g (200 mg) samples to the nearest 0.1 mg (0.0001 g). Transfer each sample to a separate flask with 10-15 ml DI and add several drops of phth 4. 5. Pipet 15 mls std. HCl into each of the six (2-CaCO,&4-Sample) flaskiExplanation / Answer
Given procedure:
1)200 mL 0.15 N HCl is prepared (using dil. 6N HCl)
2)5 mL 0.15 N HCl taken in 3 flask ( to it Di and an indicator added ….this is less important as far as your question concerned )
3)Then 5 mL 0.15 N HCl is titrated with Standardized NaOH solution and you might have taken(or will take) burette reading i.e. volume(mL) of NaOH added and as NaOH used is standardized its normality also must have given (or will be given) to you.
4)Finally they ask you to find normality of HCl using N1V1 = N2V2 (Its called as Normal equation)
You are amazed that why to find normality of HCl again as it is already known to us i.e. its 0.15 N. ok?
But what the experiment ask is correct only. To understand this let me make you aware of something.
Primary standard and standard solutions : Standard solutions are the solutions whose concentration i.e. normality or molarity do not changes for long time. These solutions are prepared from such a substances whose chemical composition do not change with time or they do not change their chemical composition for long time. They do not hydrates etc (because these processes changes concentration of solution to be formed)
And the solutions whose concentration changes with time they are termed as non-standard solutions. And dil. 6N HCl that you used for preparing 200 mL 0.15 N HCl solution is non-standard. i.e. 0.15N may not be its actual concentration.
Such non-standard solutions if needed to be used for volumetric analysis are standardized by using standar primary solutions of stanrdized solutions.
That’s what you are suppose to do with your HCl solution i.e. its standardization i.e. finding its actual concentration.
Hence you have volume of HCl in flask = V1 and N1=? (Not 0.15 N, ok?)
And for NaOH, Volume V2 = -----mL NaOH you added to HCl flask until you get faint pink colour and N2 = given to you by teacher or lab attendant.
Means you know now, N2, V2, V1 and N1 = ? you can find out by using normal equation
N1 x V1 = N2 x V2
N1 = (N2 x V2) /V1.
Note : This N1 will be actual normality of HCl solution (not 0.15 may be more or less). This N1 value only you are going to use in further calculations.
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