The equivalence point is reached when 25.00 mL of 0.100 M EDTA is titrated with 25.00 mL of 0.100 M Mg2*. Calculate the pMg2* if the titration is buffered to pH 10.00, ayt = 0.30 at that pH, and Kf = 108.79 for the MgY2- complex. O 1.00 5.05 O 7.69 O 5.43 O 9.57
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- The accompanying data (1.00-cm cells) were obtained for the spectrophotometric titration 10.00 mL of Pd(II) with 2.44 10-4 M Nitroso R(O. W Rollins and M. M. Oldham, Anal. chem .,1971, 43, 262, DOI: 10.1021/ac60297a026). Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:125.00 mL 0.01000 M Ni2+ is titrated with 0.01000 M EDTA in a solution buffered topH 5.0. Given that the formation constant for the Ni-EDTA (NiY2–) chelate is4.2 x 1018 and the 4 value at pH 5.0 is 3.54 x 10–7. Explain briefly why EDTA is an important reagent used in complexometrictitrations.50 mL of a solution of 0.0200 M Zn2+ will be titrated with 0.0100 M EDTA in 0.0100 M NH3 at pH 6.0. Ethylenediaminetetraacetic acid (EDTA) can be considered as a tetraprotic acid (H4Y). The stepwise acid dissociation constants are: K1 = 1.02 x 10-2, K2= 2.14 x 10-3, K3 = 6.92 x 10-7 and K4 = 5.50 x 10-11. The alplia value of the un-deprotonated species in a solution buffered to a certain pH is given by the following equation: a0 = [H+]4/([H+]4 + K1[H+]3 + K1K2[H+]2 + K1 K2K3[H+] + K1 K2K3K4) Calculate the alpha value of the fully deprotonated species (Y4- ) in a solution buffered to a pH of 6.0.
- Cd²+ forms a stable complex with EDTA. Consider the titration of 30.00 ml of 0.0525 M Cd with 0.125 M EDTA at pH 9, using Eriochrome black T as the indicator. Constants you may need: log K, (Cd-EDTA) 16.5. a, @ph 9= 0.052 The pCd at equivalence point (12.6 mL) is [a]A 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0418 M EDTA. The endpoint volume was 44.36 ml. A second aliquot of the same mixture was made strongly basic by the addition of NaOH – this causes the Mg2+ to precipitate as Mg(OH)2. The solution was then titrated with the 0.0418 M EDTA and the endpoint volume was found to be 30.02 ml. Calculate the molar concentration of Mg.25.00 mL 0.01000 M Ni2+ is titrated with 0.01000 M EDTA in a solution buffered to pH 5.0. Given that the formation constant for the Ni-EDTA (NiY2–) chelate is 4.2 x 1018 and the 4 value at pH 5.0 is 3.54 x 10–7 An indicator (Ind) forms a metal-indicator complex with Ni2+ (NiInd), giving aconditional formation constant of 1.00 x 108 at pH 5.0. It is generally assumedthat human eyes can detect about 1 part of one color in 10 parts of another;therefore, the first discernible color change will occur when the [NiInd]:[Ind]ratio changes from 10 to about 0.1.
- An EDTA solution was allowed to react with Pb²⁺ to produce 0.25 M PbY²⁻, 2.67×10⁻⁸ M Pb²⁺ and an excess of 0.10 M at equilibrium (K = 1.1×10⁸). What will be the α₄ value under these conditions?A 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0499 M EDTA. The endpoint volume was 40.17 ml. A second aliquot of the same mixture was made strongly basic by the addition of NaOH – this causes the Mg2+ to precipitate as Mg(OH)2. The solution was then titrated with the 0.0499 M EDTA and the endpoint volume was found to be 34.70 ml.Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 6.00. Calculate pMn21 at the following volumes of added EDTA and sketch the titration curve: 0, 20.0, 40.0, 49.0, 49.9, 50.0, 50.1, 55.0, and 60.0 mL.
- 25.00 mL 0.01000 M Ni2+ is titrated with 0.01000 M EDTA in a solution buffered topH 5.0. Given that the formation constant for the Ni-EDTA (NiY2–) chelate is4.2 x 1018 and the 4 value at pH 5.0 is 3.54 x 10–7. Calculate the pNi at the equivalence point.The sulfate in a247.1 mg sample was precipitated as BasO4 by addition of 25.00 mL of 0.03992 M BaCl2. The precipitate was removed by filtration and the remaining BaCl2 consumed 36.09 mL of 0.0217 M EDTA for titration to the Camalgite endpoint. Calculate the % SO3 in the sample.What is the equivalence volume when 0.0500 M EDTA is titrated with 100.0 mL of 0.0500 M Mn+ buffered to a pH of 9.00?
