In order to titrate EDTA into a water sample of unknown water hardness, the EDTA of known concentration is first diluted. If 25.0 mL of a 0.0632 M EDTA solution is transferred into a 250.0 mL volumetric flask, what is the concentration of the diluted EDTA solution?
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In order to titrate EDTA into a water sample of unknown water hardness, the EDTA of known concentration is first diluted.
If 25.0 mL of a 0.0632 M EDTA solution is transferred into a 250.0 mL volumetric flask, what is the concentration of the diluted EDTA solution?
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- 100 mL of mineral water containing Mg2+ and Ca2+ is taken and titrated with 86.65 mL of 0.06120 M EDTA. NH4F was added to the second 100 mL portion taken from the mineral water to mask the magnesium in the sample as MgF2, and when this sample was titrated with the same EDTA solution, 38.56 mL was spent. Find the concentration of CaCO3 and MgCO3 in the sample in ppm.You took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point. 15.52 mL of EDTA solution were used to titrate hardness that actually came from the unknown 1.552×10−4 moles of EDTA reacted with hardness-causing ions from the unknown sample 1.552×10−4 moles of hardness-causing ions were present in the unknown sample A) Assuming that the total hardness of water is due to CaCO3, how many grams CaCO3 does it correspond to? B) What is the total hardness of the unknown water in ppm CaCO3?Calculate the PCa vs. EDTA titration curve for 50.0 mL of 0.0060 M Ca2+ being titrated with 0.0150 M EDTA in a solution buffered to a constant pH of 10.00. After the addition of 0.00 mL After the addition of 5.00 mL After the addition of 26.00 mL
- The Cr plating on a surface that measured 3.00 cm x 4.00 cm was dissolved in HCl. The pH was suitably adjusted, and 10.19 mL of 0.01441 M EDTA was then introduced. The excess reagent required a 2.91-mL back-titration with 0.007171 M Cu2+. Calculate the average weight (in mg) of Cr (51.9961 g/mol) on each square centimeter of surface.b. EDTA cannot be used as a primary standard always. When can EDTA be used as a primarystandard ? Give reason why EDTA is used in complexometric titrations.The Tl in a 9.76-g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is Tl+3 + MgY-2 → TlY- + Mg+2 . Molecular mass: Tl2SO4 = 504.8 a. Titration of the liberated Mg2+ required 13.34-mL of 0.03560 N EDTA. The type of titration involved in the assay is (direct/indirect/residual) ______? b. percentage of Tl2SO4 in the sample is __________ %?
- Which of the following experiments can be described as a displacement titration? A. Calcium in powdered milk is determined by dry ashing a 1.50 g sample and then titrating the calcium with 12.1 mL of 0.008949 M EDTA. B. The Tl in a 9.57-g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is Tl3+ + MgY2- -> TlY- + Mg2+ Titration of the liberated Mg2+ required 12.77 mL of 0.03610 M EDTA. C. A 3.650-g sample containing bromide was dissolved in sufficient water to give 250.0 mL. After acidification, silver nitrate was introduced to a 25.00-mL aliquot to precipitate AgBr, which was filtered, washed, and then redissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) -> 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 26.73 mL of 0.02089 M EDTA. D. A solution contains 1.569 mg of CoSO4 (155.0 g/ mol) per milliliter. 50.00 mL of 0.007840 M EDTA…The Tl in a 9.76-g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is Tl+3 + MgY-2 → TlY- + Mg+2 Titration of the liberated Mg2+ required 15.34-mL of 0.08560 N EDTA. The type of titration involved in the assay is ___ (direct/indirect/residual) and the percentage of Tl2SO4 in the sample is ____%. Molecular mass: Tl2SO4 = 504.8A 100.0 mL sample containing Zn2+ was treated with 50.0 mL of 0.0500 M EDTA to complex all the Zn2+ and leave excess EDTA in solution. The excess EDTA was then back titrated, requiring 10.00 mL of 0.0500 M Mg2+. What was the concentration of Zn2+ in the original solution?
- You are asked to titrate a Mn3+ solution with EDTA at pH 9.00. The overall ionic strength of the solution is 0.10 M. Mn3+ +EDTA4- ⇌ MnEDTA- log K = 25.2 a. Calculate the conditional formation constant for MnEDTA- at pH 9.00 b. Calculate the equilibrium [Mn3+] at pH 9.00 for total Mn3+ = 2.0 mM i) total EDTA = 0.50 mM ii) total EDTA = 5.00 mMA standard CaCO3 solution is prepared by dissolving 0.4193g in enough dilute HCl to effect solution and then diluted to 500ml solution. A 25.00ml aliquot requires 23.62ml EDTA solution for titration. A 25.00ml water sample determined for total hardness, requires 8.45ml of the EDTA solution using Eirochrome Black T indicator. Calculate the total hardness in the water sample expressed in ppm CaCO3.Calculate the pZn2+ for solutions prepared by adding 0.00, 5.00, 10.00, 15.00, 20.00, 25.00 and 30.00 mL of 0.0100 M EDTA to 25.00 mL of 0.00250 M Zn2+. Assume that both the Zn2+ and EDTA are 0.0100 M in NH3 to provide a constant pH of 9.0