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- C1. A carefully weighed 280mg Calcium carbonate was used in the standardization of an EDTA solution. Initially, 25mL of the titrant was consumed but only after the addition of another 10mL of the titrant did an endpoint was visible. What is the molar concentration of the standardized EDTA solution? (use C1 as reference)* 0.08m 0.8M 0.08N 0.08M How many grams of EDTA (MW: 292 g/mole) is required to prepare 250mL of a 0.025M solution?* 0.1825 0.1852 1.825 1.1852A 50.00-mL aliquot of solution containing 0.310 of MgSO4 (FM 120.37) in 0.500 L required 37.77 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution? Answer in 3 significant figures. Do not include the unit. need asap0.4545 g CaCO3 was dissolved in HCl and the resulting solution was diluted to 0.25 L. Twenty-five mL aliquot of this solution required 35.2 mL of EDTA upon performing titrimetric analysis. Determine the mass of NaH2Y·2H2O to prepare 0.5 L titrant solution. NaH2Y·2H2O is a cmpound with MW = 372.2 g/mol Note: H2Y2- = EDTA
- A CaCO3 solution that will be used to standardize EDTA was prepared by dissolving 3.1251 g of solid CaCO3 in 100 mL dilute HCl. A 20.00 mL aliquot was taken for titration with EDTA consuming 32.00 mL of the titrant to reach the endpoint. Express the concentration of EDTA in molarity and in CaCO3 titer. MM Na2H2Y2•2H2O = 372.24 g/mole; MM CaCO3 = 100.09 g/mole1. How much volume (mL) is needed to standardize 0.05M of EDTA using 180mg of magnesium ribbon? MW: 24g/mol2. How many grams of EDTA is required to prepare 250mL of a 0.05M solution? MW: 292g/mol3. What is the Molarity of the EDTA solution if 35mL of the titrant was added to 0.2g of calcium carbonate? Round off to four decimal placesA 46.4046.40 mL aliquot from a 0.4850.485 L solution that contains 0.3500.350 g of MnSO4MnSO4 (MW=151.00MW=151.00 g/mol) required 35.235.2 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO3CaCO3 ( MW=100.09 MW=100.09 g/mol) will react with 1.691.69 mL of the EDTA solution?
- 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 = 504.8Titration of 25.00ml of 0.100M I- with 0.0500M Ag+. Calculate pAg+ (=-log[Ag+]) at VAg+= 10.00ml; 49.50ml; 50.00ml; 50.50ml; 52.00ml. (Ksp,Ag+)= 8.3*10^-17Commercial Vanadyl Sulfate (VOSO4) is contaminated with H2SO4 and H2O. A solution was prepared dissolving 0.2447 g of impure VOSO4 in 50.00 mL of water. A spectrophotometric analysis indicated that the The concentration of vanadyl ions, VO2+ (blue) was 0.0243 M. A 5.00 mL sample was passed through a column filled with a cation exchange resin in H+ form, to retain the vanadyl ion and the eluent necessary for its titration 13.03 mL of NaOH 0.02274 M. Calculate the percentage by weight of each component (VOSO4, H2SO4, and H2O) in commercial vanadyl sulfate. (Pm VOSO4, 162.96; H2SO4, 97.94)
- 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 60.00 mL of buffered solution (pH 10) containing Zn and Nit ions (from a chip sample) is analyzed using EDTA titration with EBT as indicator. The following data were recorded: M of EDTA = 0.05204 V of EDTA for Zn2+ =D41.97 V of EDTA for Ni2+ mL = 15.16 mL How many mmoles of Zn and Ni were present in the solution? If the weight of the chip is 1.000 g, give the % Zn and % Ni in the original sample.A solution containing 1.263 g of unknown potassium compound was dissolved in water and treated with excess sodium tetraphenylborate, Na1B1C6H524 solution to precipitate 1.003 g of insoluble K1B1C6H524 (FM 358.33). Find wt% K in the unknown.