What is the concentration of cobalt in a solution that contained: 15 mL of cobalt 7.2...
What is the concentration of cobalt in a solution that contained:
15 mL of cobalt
7.2 mL of 50% m/v NH4SCN
30.94 mL of Acetone
This solution was then completely titrated with 1.79 mL of 0.1 M of EDTA
Homework Answers
so the concentration in the
given smple is 4.8 mg / 15 mL = 0.32 mg/mLAdd Answer to:
What is the concentration of cobalt in a solution that
contained:
15 mL of cobalt
7.2...
The metal ion Mn+ was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL...
The metal ion Mn+ was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (Mn+) buffered at a pH of 9.00. a) Calculate the equivalence volume (Veq). b) What is the concentration of the free metal ion at volume V = Veq/2. c) If the conditional formation constant, K’f = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Veq.
What is the concentration of a 53.80 mL solution of HBr that is completely titrated by...
What is the concentration of a 53.80 mL solution of HBr that is completely titrated by 27.5 mL of a 0.200 M NaOH solution?
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point (i.e. 50.0 mL EDTA has been added), what is the equilibrium concentration of Ca2, [Ca2...
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point (i.e. 50.0 mL EDTA has been added), what is the equilibrium concentration of Ca2, [Ca2], and what is the pCa? The formation constant of CaY, Kaa 5.0 x 1010, and α4 of EDTA at pH 10.0 is 0.35
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point...
What is the Co2+ concentration after 50 mL of a 0.030 M Co2+ solution is mixed...
What is the Co2+ concentration after 50 mL of a 0.030 M Co2+ solution is mixed with 50 mL of a 0.050 M EDTA solution at pH 10?
A 25.0 mL aliquot of 0.0440 M EDTA was added to a 49.0 mL solution containing...
A 25.0 mL aliquot of 0.0440 M EDTA was added to a 49.0 mL solution containing an unknown concentration of V3+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back-titrated with a 0.0340 M Ga3+ solution until all of the EDTA reacted, requiring 11.0 mL of the Ga3+ solution. What was the original concentration of the V3+ solution?
10. Assume you are carrying out the titration of 50 mL of a 0.025 M solution...
10. Assume you are carrying out the titration of 50 mL of a 0.025 M solution of acetic acid with 0.1023 M NaOH. Acetic acid has a Ka of 1.8 × 10−5 . (a) Calculate the pH of the solution at V = 0, V = 0.3Veq, V = Veq, and V = 1.2Veq. (Veq is the equivalence point volume) (b) If a phenolphthalein indicator was used in this titration, where would the apparent endpoint occur? Assume the apparent endpoint...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of the unknown was treated with 25.00 mL of 0.2404 M EDTA. The excess EDTA was then back titrated with 8.52 mL of 0.0694 M Zn2+ to reach the equivalence point. What was the concentration of Ni (in unit of M) in the 50.00 mL sample? Please keep your answer to three decimal places.
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL...
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+-BAL complex. The excess Pb2+ was titrated with 0.0157 M EDTA, requiring 9.88 mL to reach the equivalence point. Separately, 39.89 mL of the EDTA solution was required to titrate 32.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution concentration:
A 20.21 mL aliquot of a Pb2+ solution, containing excess Pb2+, was added to 11.50 mL...
A 20.21 mL aliquot of a Pb2+ solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+ –BAL complex. The excess Pb2+ was titrated with 0.0136 M EDTA, requiring 8.21 mL to reach the equivalence point. Separately, 40.35 mL of the EDTA solution was required to titrate 31.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution.
(10) (3) 4. The metal ion Mnt was titrated with 0.0500 M EDTA. The initial solution...
(10) (3) 4. The metal ion Mnt was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (M9+) buffered at a pH of 9.00. a) Calculate the equivalence volume (Veg). b) What is the concentration of the free metal ion at volume V = Veg/2. c) If the conditional formation constant, K'r = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Veq. (4)
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point (i.e. 50.0 mL EDTA has been added), what is the equilibrium concentration of Ca2, [Ca2], and what is the pCa? The formation constant of CaY, Kaa 5.0 x 1010, and α4 of EDTA at pH 10.0 is 0.35
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point...
What is the Co2+ concentration after 50 mL of a 0.030 M Co2+ solution is mixed with 50 mL of a 0.050 M EDTA solution at pH 10?
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+-BAL complex. The excess Pb2+ was titrated with 0.0157 M EDTA, requiring 9.88 mL to reach the equivalence point. Separately, 39.89 mL of the EDTA solution was required to titrate 32.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution concentration:
(10) (3) 4. The metal ion Mnt was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (M9+) buffered at a pH of 9.00. a) Calculate the equivalence volume (Veg). b) What is the concentration of the free metal ion at volume V = Veg/2. c) If the conditional formation constant, K'r = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Veq. (4)
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