Question

Please help with my Titration lab! I am having difficulty understanding this. Thank you!

C. Analyzing an unknown base - optimizing a method: In this segment you will design your own method for analyzing an unknown base. In the solutions list there are several different concentrations of HCl to choose from. You can also try different sample volumes of the unknown base. Be sure to record these for each trial. Your goal is to find two different combinations of HCI titrant and sample volume of base that give satisfactory results. A "good result should require 5 to 49 mL of the HCI titrant (so not less than 5 mL, and more than 50 mL is not usable). 1. Add the following to your workbench: a 250 ml Erlenmeyer, a pipette, the unknown base, and bromocresol green indicator 2. Transfer some of the unknown base to the flask using the pipette record the volumel, and add 040 ml of bromocresol green from the dropper bottle. This indicator is blue in basic solution and yellow in acidic solution, so your sample will initially be blue. 3. Remove the solutions and pipette from your workspace Add a 50 mL burette and one of the HCl solutions (under strong acids" in "solutions) Add 520 ml of your chosen HCI titrant to your burette, and record its molarity 4 Drag the burette over the sample flask to set up your titration Record the initial meniscus reading 1000 ml) Enter 0.10 mL in the volume box for the burette. Click the "pour" button until the solution turns yellow (dont click too fast-the blue will fade to green, and then one click will cause the change to yellow). Record the final burette reading. If you run out of titrant, list">50.00 mL for your final reading. 5. Perform at least three additional trials, but change either the HCL titrant concentration or the sample volume of unknown base. Think about your results and consider whether using more or less concentrated titrant for a smaller or larger volume of unknown) may affect the next titration. 6. You must obtain at least two trials with usable final readings that can be analyzed (ie, not >50.00 mL). If needed perform additional trials and record the data in the second table below. Table 3: Titration of Unknown Base Trial 1 Trial 3 Trial 4 Trial 2 25.4 Volume of unknown base 30.0 25.4 юм Titrant (HCI) molarity Initial burette reading 0.0 30.0 COM 4.0 4,7 0.7 016 Final burette reading Volume of acid used 3.9 3.1 118 (for additional trials, only if needed): Trials Trial 6 Trial 7 Trial 8 Volume of unknown base Titrant (HCI) molarity Initial burette reading Final burette reading Volume of scid used

Answer 1

se :N, U = NzV₂ N, 2 Normality of Hel VE Volume of hel.. Ni: Normality of base U₂= volume of base -> for neutralizohen si eaction to be complete, number of gram equivalent should be equal # of gram equivalent Nov N = Normality of solution ve volume of solution am # 07 .-) as for your case, base has only . one hydrosy group iso Normality. is equal to Molarity. A generally. N = n. m ( for base). N normality na no. of hydow'de group Mo Molarity base has a Scanned with Cor valency factor ) CamScanner K URSUS

Trial to : N. VI = NzV2 е не ибжола ? Че (а valoncy factor (n) is equal to t because her give one get - 10x 0.6 25.4 x 1 Ny 0.236 N. Ľ M₂ = 0.236M -) Molasity - Normality g base ? 0:2:36 Trial 2: Ng Vi = N₂V₂ 3x1.8= N2 x 25.4 I Nz 20-2126 N 7. I M2 = 0.2126M - Molasity: Normality y base ? 0:2126 Scanned with CS CamScanner

Trial 3: Nivia NzV₂ 10x 9.73 30X NZ I M = 0.233N7 [ M2 = 0+23314 * Normality Molasity og base a 0:233 Trial 4 NIN, a Navz .3x2 1 2 NqX3.0 I N.22 0.21N 7 L M₂2 0.21M 1 a Normality = Molarity of base = 0.21 to for first question, you should go for the trial 2" and "Trial 4.", because volume of He used is higher than other trial and the higher volume. help us in minimizing the error CS Scanned with CamScanner

1.2. Let we go for Trial ? and Trial 4. In both trial, volume of base= 30 mL and Hel concentrations are different :lom a 3M. a as number of oh ion in the solution are same for both trials, so i we need same number of het nous for neutralization , is but contentration of HCl we used are different so, we need different volume of fill in corresponding Trials, - Af we consider that, then between volume will be, relationship . 10m x V₂ = ? M X Vu 4 : : V22 volume of Hel in Trial-3 Vu= volume of Hel in Trialny ca Scanned with CamScanner

3. Now, again NVIE N₂U₂ is for tilli normality WOX is equal to molarity 0.1 X 46.55 2. N2X12.5 N2 = 0.3724N Normality of base = 0.3724 N but here the buse boas horee hyshony group, so, valency factor will be (n=3 ] Now. Molarity - Normality . Valanly factor Molasity 0.3.724 . & Molarity = 0.124 M 7 les scanned With Mola sity of base ... CamScanner