Table 1. Concentration of acetic acid in the vinegar sample Concentration of Volume of acetic acid Concentration of acetic acetic acid (mL) acid (diluted) (undiluted) Trial 1 37.19 Trial 2 36.83 Trial 3 37.04 Average Note: Concentration of standardized NaOH = 1.100 M Equation to obtain the undiluted concentration: CV1 = C;V2 C1 = concentration of acetic acid (undiluted) C2 = concentration of acetic acid (diluted) V1 = 25.00 cm³ or 25.00 ml volume taken from the vinegar sample V2 = total volume (volume of volumetric flask) %3D %3D

Table 1. Concentration of acetic acid in the vinegar sample Concentration of Volume of acetic acid Concentration of acetic acetic acid (mL) acid (diluted) (undiluted) Trial 1 37.19 Trial 2 36.83 Trial 3 37.04 Average Note: Concentration of standardized NaOH = 1.100 M Equation to obtain the undiluted concentration: CV1 = C;V2 C1 = concentration of acetic acid (undiluted) C2 = concentration of acetic acid (diluted) V1 = 25.00 cm³ or 25.00 ml volume taken from the vinegar sample V2 = total volume (volume of volumetric flask) %3D %3D

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 121AP: Calcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion...

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Calcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion are usually prepared by dissolving calcium carbonate in acid. What mass of CaCO3 should be taken to prepare 500. mL of 0.0200 M calcium ion solution?
made upetement Zazum Adomica Abmcmas rop 418.28 Specific heat 19.42 173 13
DATA AND CALCULATIONS Show all calculations neatly on an attached sheet. Trial 1 Trial 2 Trial 3 70.6849. 68.0749 2.61g 23.25ML O.65ML 22.6mL Mass of flask + vinegar Mass of empty flask Mass of vinegar used Final buret reading Initial buret reading Volume of NaOH used Moles of NaOH used 0.00226 mol Moles of açetic acid titrated Mass of acetic acid titrated Mass % acetic acid in vinegar 5.0%
3. Given: 1.000 mL N2OH E 0.0302 g H2C2O4 .2H20 ; 1.000 mL HCI E 0.1123 g BaCO3. What is the value of each mL of the HCl in terms of mL of NaOH ?
An aqueous solution is made of the diprotic acid H2A to a concentration of 0.15 M H2A. Ka1 = 3.0 x 10-4 and Ka2 = 2.5 x 10-9 What is the [A2-] in this solution? Your answer should have 2 significant figures and be expressed in scientific notation (i.e. 3.0 x 10-2 as 3.0e-2)
CONCENTRATION DETERMINATION Buret contains NaOH solution *Sodium hydroxide solution: To 1 part reagent grade NaOH add 1 part distilled, carbon dioxide-free water by weight Erlenmeyer flask contains about 9ml of the acid and about 10 ml of distilled water made recently free from CO2 by boiling. DATA SHEET II Final Reading HCI Initial Reading HCI 9.4 Final Reading NaOH Initial Reading NaOH 10.65 10.55 1 ml HCI 1 mL NAOH mL NAOH mL NaOH mL HCL mL HCL Mean Values: 1 ml HCI = 1 mL NaOH = _mL NaOH mL HCL COMPUTATION:
3 Solid ammonium bromide is slowly added to 75.0 mL of a 0.283 M silver nitrate solution until the concentration of bromide ion is 0.0583 M. The percent of silver ion remaining in solution is %. Submit Answer $ 4 R F % 5 Use the References to access important values if needed for this question. Retry Entire Group 5 more group attempts remaining T G Cengage Learning | Cengage Technical Support ^ 6 MacBook Pro Y H & 7 U J * 00 8 1 ( 9 K O < ) O L P Previous Email Instructor + = Next Save and Exit
3 Solid potassium fluoride is slowly added to 75.0 mL of a 0.233 M calcium bromide solution until the concentration of %. fluoride ion is 0.0632 M. The percent of calcium ion remaining in solution is Submit Answer 20 E $ 4 R F Use the References to access important values if needed for this question. Retry Entire Group % 5 Cengage Learning Cengage Technical Support T G ^ 6 6 more group attempts remaining B Y H ← & 7 U N * 00 8 J ▶ M ( 9 K O ) 0 L P Previous Email Instructor + 11 [ Next> Save and Exit 11 1 d
ml?ClassID=1471954077 44.00 mL of 0.50 M H2SO4 reacts with 0.35 M LIOH and reaches the endpoint using bromthymol blue as an indicator. What volume of LIOH is required for this reaction? 2LIOH + H2SO4 → LI¿SO4 + 2H2O [?] mL LIOH mL LIOH Enter
You are asked to weigh 2 g of lithium and add it to 1500 ml of distilled water +2drops of indicator, knowing that the density of water is 1 at room temperaturewhile conducting this experiment. After calculating the amount of LiOH, you need to dilute to prepare 0.25 M of LiOH in a 3000 ml solution. Show your calculation and units.Knowing that the initial volume that you used was 1500 ml. Use this equation to calculate ( CLiOH x VLiOH) before dilution = (CLiOH x VLiOH) after dilution
If 2 ml of 25 N H2SO4 is transferred to 50 ml volumetric flask and complete to mark with ditilled water, the concentration of acid will be .. M* O 2.5 O 1.5 2 O 0.5 0 1 Non of these
Calculate the percentage CH3COOH in a sample of vinegar from the following data.Sample = 15.00 g, NaOH used = 43.00 ml ; 0.600 N H2SO4 used for back titration =0.250 ml ; 1.00ml NaOH is equivalent to 0.0315 g H2C2O4. 2H2O.