Question

A 0.4505 g sample of CaCO3 was dissolved in the HCl and the resulting solution diluted...

A 0.4505 g sample of CaCO3 was dissolved in the HCl and the resulting solution diluted to 25.00 mL in a volumetric flask. A 25.00 mL aliquot of the solution required 29.25 mL of EDTA solution for titration to the Eriochrom Black T end point.

a. How many moles of CaCO3 were present in the solid sample?

b. What is the molar concentration of Ca^2+ are contained in a 250.00 mL aliquot of the CaCl2 solution?

c. How many moles Ca^2+ are contained in the 25.00 mL aliquot of CaCl2 solution?

d. How many moles EDTA are contained in the 29.25 mL used for the titration?

e. What is the molar concentration of the EDTA solution? f. From the molarity of the calcium solution you found in part (b), calculate the concentration in ppm of calcium ion and the concentration in ppm of CaCO3

Homework Answers

Answer #1

a.moles CaCO3 = 0.4505 g/ 100.1 g/mol=0.0045

b.[Ca2+]= 0.0045/ 0.250 L=0.018M
c.moles Ca2+ = 0.018 x 0.02500 L=0.00045

d.moles of EDTA can be found from its concentration using the formula = M1V1=M2V2

[Ca2+]0.018M*25 = M2*29.25 [EDTA]

[EDTA=]M2=0.0153M

moles of EDTA = 0.0153M*0.02925 = 0.000447

e) Molarity of EDTA = 0.0153M

f) [Ca2+]= 0.0045/ 0.250 L=0.018M

0.018 mol/L times 40.08 g/mol = 0.721g/L

0.721 g / 1000 g of solution times 1000/1000 = 400 g / 1,000,000 g of solution

The answer is 721 ppm

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A 0.3205g sample of CaCO3 was dissolved in HCl and the resulting solution diluted to 250.ml...
A 0.3205g sample of CaCO3 was dissolved in HCl and the resulting solution diluted to 250.ml in a volumetric flask. A 25 mL sample of the solution required 18.75mL of EDTA solution for titration to the Eriochrome Black T end point. What is the concentration of Ca2+ (mol/L) in the 250.0ml of CaCl2 solution? How many mole of Ca2+ are contained in a 250mL sample? How many mole of EDTA are contained in the 18.75mL used for titration? What is...
A 0.4858 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid....
A 0.4858 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2·4H2O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 200.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.85 mL of 0.001523 M EDTA. Thiosulfate was used to mask the copper in a second 20.00 mL aliquot. Titration of...
An unknown amount of a compound with a molecular mass of 259.61 g/mol is dissolved in...
An unknown amount of a compound with a molecular mass of 259.61 g/mol is dissolved in a 10-mL volumetric flask. A 1.00-mL aliquot of this solution is transferred to a 25-mL volumetric flask and enough water is added to dilute to the mark. The absorbance of this diluted solution at 347 nm is 0.525 in a 1.000-cm cuvette. The molar absorptivity for this compound at 347 nm is ε347 = 6557 M–1 cm–1. (a) What is the concentration of the...
A 1.5000-g sample of an alloy was dissolved in acid and the tin was precipitated as...
A 1.5000-g sample of an alloy was dissolved in acid and the tin was precipitated as H2SnO3 · xH2O.  After filtration and ignition, the resulting precipitate of SnO2 weighed 0.1430 g.  The filtrate was diluted to 1000.0 mL.  A 20.00-mL aliquot was adjusted to the appropriate pH and required 33.18 mL of 0.01047 M EDTA for titration of the Cu and Pb.  A 25.00-mL aliquot of the filtrate was treated with Na2S2O3 to mask the Cu and was then titrated...
1.5 g of caco3 (M.W 100.09g/mol) are transferred to a 500 ml volumetric flask and 1:1...
1.5 g of caco3 (M.W 100.09g/mol) are transferred to a 500 ml volumetric flask and 1:1 Hcl is added dropwise until effervescence ceases and the solution is clear. everything is diluted with water in the mark. 30 ml of that solution are titrated with EDTA and 15.5 ml are spent to reach the endpoint. Calculate the molarity of EDTA
Information given: The following data is recorded, For the following KHP solution, 10.575g of KHP is...
Information given: The following data is recorded, For the following KHP solution, 10.575g of KHP is dissolved water, and diluted to 250.00mL in a volumetric flask; 20.00mL of the KHP solution are added to an Erlenmeyer flask; a buret is filled with NaOH solution of unknown concentration to a starting volume of 0.18mL, and at the end point if the titration the buret reading is 20.83mL. 1.) Calculate the moles of KHP that were put into the 250.00mL volumetric flask....
20.40g Iron(III) oxalate is dissolved in enough water to give 250.0 ml of solution. 25.00 ml...
20.40g Iron(III) oxalate is dissolved in enough water to give 250.0 ml of solution. 25.00 ml of this solution is pipetted into a 100.0ml volumetric flask and diluted to the mark. A. Calculate the molarity of Iron(III) oxalate in the original solution. B. Calculate the molarity of the oxalate ion in the diluted solution.
A student pipetted 25.00 mL of a stock solution that was 0.1063 M HCl into a...
A student pipetted 25.00 mL of a stock solution that was 0.1063 M HCl into a 100.00 mL volumetric flask and diluted the solution to the volumetric flask calibration mark with deionized water. Calculate the concentration of the diluted solution.
A sample of paint chips (0.6169g) was added to 10 mL of concentrated HNO3 and heated...
A sample of paint chips (0.6169g) was added to 10 mL of concentrated HNO3 and heated to near dryness. Approximately 35mL of water was added and the solution was biolded for 30 mins. After cooling, the solution was filtered and the filrate was diluted to volume in a 50mL volumetric flask. A 5 mL aliquot of this solution was diluted to volume in a 25mL volumetric flask. The lead concentration of the final diluted solution (in the 25mL volumetric flask)...
A 3.826 g sample containing Fe, Ti, and other inert material was dissolved and diluted to...
A 3.826 g sample containing Fe, Ti, and other inert material was dissolved and diluted to form 250.0 mL of solution. The dissolution conditions were such that Fe was converted to Fe(III) and Ti was converted to Ti(IV) in solution. A 25.00-mL aliquot of the solution was then passed through a Walden reductor and titrated with 0.2032 M Ce4 . The endpoint was reached following the addition of 22.42 mL of the Ce4 solution. A second 50.00-mL aliquot of the...