Molarity of EDTA .005547A Unknown number 7342 Source of Water Come Park Part A V

Question

Molarity of EDTA .005547A Unknown number 7342 Source of Water Come Park Part A Volume of unknown used 2Snl Volume of hard water sample 5Onl (Place a star next to the trials you are using.) Trial 1 Trial 3 Trial 4 if needed) (if needed) Trial 2 Trial 5 Initial buret reading lnl Snt. Final buret reading | 193AL 121.8m 20.1AL 20.GML Volume tirant added 1 18.9 AL ML 9.3 nL 18& L LG- Part B (Place a star next to the trials you are using.) Trial 1 Trial 4 (if needed)(if needed) Trial 2 Trial 3 Trial 5 Initial buret reading 12 12mL-11ml Eimal buret readingS 5.LIS. I n Volume titrant added Observations before, during, and after the titration: Part A l. For the following calculations, use the three trials that were within 0.50 mL. Calculate the moles of EDTA used to reach the endpoint for each trial. Trial #2: Trial # ? Trial # o05547 x 13. ,005547 x13.2 SOAL ,0015 M .005347 x13. 56AL 00Is m 50nl .0015 m Expt 9-Page 7

Explanation / Answer

Part A)

(1) Trial 2

moles EDTA used = 0.005547 M x 18.8 ml = 0.1043 mmol

Trial 3

moles EDTA used = 0.005547 M x 18.9 ml = 0.105 mmol

Trial 4

moles EDTA used = 0.005547 M x 18.6 ml = 0.1032 mmol

(2) moles Ca2+ present = moles EDTA used

Trial 2

moles Ca2+ present = 0.1043 mmol

Trial 3

moles Ca2+ present = 0.105 mmol

Trial 4

moles Ca2+ present = 0.1032 mmol

(3) original molarity of Ca2+

Trial 2

[Ca2+] present = 0.1043 mmol/25 ml = 0.0042 M

Trial 3

[Ca2+] present = 0.105 mmol/25 ml = 0.0042 M

Trial 4

[Ca2+] present = 0.1032 mmol/25 ml = 0.00413 M

average molarity = 0.0042 M

(4) g CaCO3 in 1.000 L water

= 0.0042 M x 100.1 g/mol

= 0.420 g/L

(5) In mg/L

= 0.420 g/L x 1000

= 420 mg/L

Is the hardness of water

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Part B)

(1) Trial 2

moles EDTA used = 0.005547 M x 13.1 ml = 0.0727 mmol

Trial 3

moles EDTA used = 0.005547 M x 13.2 ml = 0.0732 mmol

Trial 4

moles EDTA used = 0.005547 M x 13.1 ml = 0.0727 mmol

(2) moles Ca2+ present = moles EDTA used

Trial 2

moles Ca2+ present = 0.0727 mmol

Trial 3

moles Ca2+ present = 0.0732 mmol

Trial 4

moles Ca2+ present = 0.0727 mmol

(3) original molarity of Ca2+

Trial 2

[Ca2+] present = 0.0727 mmol/50 ml = 0.00145 M

Trial 3

[Ca2+] present = 0.0732 mmol/50 ml = 0.0015 M

Trial 4

[Ca2+] present = 0.0727 mmol/50 ml = 0.00145 M

average molarity = 0.0015 M

(4) g CaCO3 in 1.000 L water

= 0.0015 M x 100.1 g/mol

= 0.150 g/L

(5) In mg/L

= 0.150 g/L x 1000

= 150 mg/L

Is the hardness of water

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5. The hardness in Part A) is much higher than the hardness in part B found after treatment.

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