Concentration of HCl used: 25ML Volume of HCl used: 0.1031 Indicator: phenaiphthalein (pneno) Endpoint colour change: Light pink /rea Table 1: Titration of Sodium Hydroxide with standard Hydrochloric Acid. Burette reading (mL) Trial 2 Trial 3 Trial 4 Trial 1 (if required) Final reading Initial reading Volume of NaOH(aq) 26.3 mL O. 00mL 27:2mL O.00mL 2니 mL 0.00mL 24mL 26.3mL 27.2mL used Average of 2 closest values to within 0.10 mL: 26.75ML 1. Write the molecular, total and net ionic equations for the neutralization reaction. + HCl at NaOH caa)→ Naci capit H2O n → Nat + C1- t H20 H+ CI t Nat + OH- H*caq) + OH (aq) → HzO() 51 A I Iydroxide in

Concentration of HCl used: 25ML Volume of HCl used: 0.1031 Indicator: phenaiphthalein (pneno) Endpoint colour change: Light pink /rea Table 1: Titration of Sodium Hydroxide with standard Hydrochloric Acid. Burette reading (mL) Trial 2 Trial 3 Trial 4 Trial 1 (if required) Final reading Initial reading Volume of NaOH(aq) 26.3 mL O. 00mL 27:2mL O.00mL 2니 mL 0.00mL 24mL 26.3mL 27.2mL used Average of 2 closest values to within 0.10 mL: 26.75ML 1. Write the molecular, total and net ionic equations for the neutralization reaction. + HCl at NaOH caa)→ Naci capit H2O n → Nat + C1- t H20 H+ CI t Nat + OH- H*caq) + OH (aq) → HzO() 51 A I Iydroxide in

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.19QAP

See similar textbooks

Similar questions

ALL INFORMATION NEEDED IS PROVIDED, I NEED FIRST QUESTIONS ANSWERED FIRST BEFORE OTHERS CAN BE ANSWEREDRate = -(Delta[RCl])/Delta t andln[RCl]t=-k*t+ln[RCl]0 < [R-Cl]-time equation Test tube A = 0.30 of 0.10M NaOH(aq), 6.70ml DI H2O & 1 drop of bromophenol bue Test Tube B = 3.00M of 0.10M R-CL in acetone. Mix together in 3 water baths Bath A Trial 1= 20C, 53.6s Trial 2= 20C, 49s for color to changeBath B Trial 1= 32C, 18s Trail 2= 32C, 20s Bath C Trial 1= 10C, 204s Trial B= 10C, 200s average out both trials for each bathcolor change happens after 10% of R-Cl has reacted so In terms of R-Cl this means that 10% of the R-Cl has reacted and ____ of the R-Cl is left Q1. at the time of the color change [R-Cl]t=____[R-Cl]o (Fill in the blank) Q4. Substitute the value of [R-Cl]t into appropriate equation listed above and solve for k in terms of [R-Cl] and t. Q5. Use average time for each temp and determine the value of k (with units) at each temperature Q6. Rearrange the Arrhenius…
What is the percentage purity of acetic acid if 2.6 grams required 32.5 ml of 0.994 N NaOH solution to reach the endpoint? Does it conform to USP requirements for acetic acid 3%-6%?
why is it necessary to test the ph? PROCEDURE To a six inch test tube was added thiamine hydrochloride (0.65 g), followed by water (2mL). The mixture was gently shaken until homogeneous, and then ethanol (7.5 mL)was added. Finally, sodium hydroxide (10%, 1.5 mL) was added and the mixture wasgently stirred to ensure complete mixing.Benzaldehyde (3.8 mL) was then added at room temperature in one portion with gentlestirring. The pH of the resulting reaction mixture was determined and adjusted to >10by dropwise addition of 10% sodium hydroxide.
The ethyl acetate (CH3COOC2H5) concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL aliquot of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH: ethyl acetate.JPG After cooling, the excess KOH was back-titrated with 3.85 mL of 0.04644 M H2SO4. Calculate the %(w/v) CH3COOC2H5 in the alcoholic solution. MM CH3COOC2H5: 88.11 MM NaOH: 40.00 MM H2SO4: 98.08
Objective: To determine the percent NaCl (w/v) of a water sample 5.00 mL sample diluted to 50 mL solution 0.1000 M AgNO3 Volume of AgNO3 trial 1=15.30 mL trial 2=14.90 mL trial 3=15.10 ml
Tim is interested in analyzing for Pyraclostrobin (pesticide) in strawberries using QuECHers method for extraction and clean up before analysis with HPLC-uv. The concentration of the Pyraclostrobin in the fruits that he sampled is thought to be between 1-10 ng pesticide per g fruit fresh weight. The HPLC-uv instrument he is using has a LoQ of 5 ppb for Pyraclostrobin. Tim is starting out with 500 mg of the sliced fruit
The buret was filled with 0.100 M HCl solution. Then was transferred in a 25.0 mL of saturated calcium hydroxide solution (2g of calcium hydroxide per 100 ml of water) in two separate E-flasks. Then 2 drops of phenolphthalein was added to each flask Titration data for the determination of solubility and Ksp of calcium hydroxide: Trial 2: Final Buret reading (ml)-26.10; Initial Buret reading (ml)- 19.80; Temperature (Celcius)- 25 Voume of HCl used: Trial 2- 6.30mL 1. Compute for the moles of H+ used and the moles of OH- present. moles of H+ used = (concentration of HCl) × (volume of HCl used)moles of OH- = moles of H+ used 2. Construct an ICE table for the reaction.3. Calculate the molar solubility (in mol/L) of OH- and Ca2+.4. Determine the solubility of Ca(OH)2 in g/L. (MM of Ca(OH)2 = 74.096 g/mol). 5. Calculate the Ksp of Ca(OH)26. Compute for the percent error of the experimental value for Ksp of Ca(OH)2 with the literature…
The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH:CH3COOC2H5 + OH- → CH3COO- + C2H5OHAfter cooling, the excess OH2 was back-titrated with 3.41 mL of 0.05042 M H2SO4. Calculate theamount of ethyl acetate (88.11 g/mol) in the original sample in grams
1. Why is there a need to correct conductivity of the sample solution?2. Cite three possible sources of error in the experiment and discuss their effects on the computed ionization constant of the acetic acid. (See details below----- literature value- Ki, acetic acid= 1.75 x 10^-5). 3. What is the effect of dilution on the fraction of the acetic acid ionized?
The buret was filled with 0.100 M HCl solution. Then was transferred in a 25.0 mL of saturated calcium hydroxide solution (2g of calcium hydroxide per 100 ml of water) in two separate E-flasks. Then 2 drops of phenolphthalein was added to each flask Titration data for the determination of solubility and Ksp of calcium hydroxide: Trial 1: Final Buret reading (ml)-19.80; Initial Buret reading (ml)- 13.00; Temperature (Celcius)- 25 Trial 2: Final Buret reading (ml)-26.10; Initial Buret reading (ml)- 19.80; Temperature (Celcius)- 25 Voume of HCl used: Trial 1- 6.80ml; Trial 2- 6.30mL 1. Compute for the moles of H+ used and the moles of OH- present. moles of H+ used = (concentration of HCl) × (volume of HCl used)moles of OH- = moles of H+ used 2. Construct an ICE table for the reaction.3. Calculate the molar solubility (in mol/L) of OH- and Ca2+.4. Determine the solubility of Ca(OH)2 in g/L. (MM of Ca(OH)2 = 74.096 g/mol). 5.…
A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.569 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Please calculate the concentration of lead in the sample in ppm.
Following the monograph procedure, a 724-mg of aspirin (MW=180 g/mol) dissolved in 18.5 ml of cold neutralized alcohol. This solution was then initially titrated with 0.101 N sodium hydroxide solution, then later neutralized with 0.104 sulfuric acid. What is the percentage purity of the sample? 1. What is the milliequivalent weight consumed by the acidic titrant? a. 1.5392 g-meq b. 2.0907 g-meq c. 1.4948 g-meq d. 2.1528 g-meq 2. What is the milliequivalent weight consumed by the basic titrant? a. 5.8656 g-meq b. 1.5392 g-meq c. 5.6964 g-meq d. 1.4948 g-meq 3. What is the difference of milliequivalent weight consumed in the reaction? a. -4.1572 g-meq b. 4.3708 g-meq c. 0.5515 g-meq d. 4.1572 g-meq