If 200 ml of 0.01 M EDTA is added to 100 ml of 0.01 M Ni solution that is buffered at PH of 10.2, The PNi of the solution will be: ( KMY= 1.87 x101) А. 2 В. 2.48 С. 10.3 D. 18.27
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The answer is D, do you know how please?
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- In an unprecedented initiative, the professor chose you to carry out the determination of a special sample in order to verify the inherent indication system. For that, an aliquot of 25.00 mL of a solution containing Fe(III) salts was titrated with EDTA 0.00982 mol/L, using potassium thiocyanate as indicator. In this case, you have found the volume of 31.10 mL of titrant to End Point. Based on the information given in the table, choose the option that best describes/explains the process linked to the indication system of the Final Point of the degree: (a) The indication of the system will be given by the disappearance of the reddish coloration, due to the displacement reaction: Fe(III) complex with EDTA is formed in detriment of the Fe(III) complex with thiocyanate. (b) The indication of the system will be given by the appearance of a reddish color, in function of the displacement reaction: Fe(III) complex with thiocyanate is formed in detriment of the Fe(III) complex with EDTA. (c) The…The concentration of a solution of EDTA was determined by standardizing against a solution of Ca2+ prepared from the primary standard CaCO3. A 0.4025 g sample of CaCO3 was transferred into a 250 mL Erlenmeyer flask and the pH adjusted by adding 5 mL of a pH 10 NH3-NH4Cl buffer containing a small amount of Mg2+EDTA. After adding calmagite as a visual indicator, the solution was titrated with the EDTA requiring 42.36 mL to reach the end point. Report the molar concentration of the titrant.A 200.00 mL solution of 0.00105 M AB4 is added to a 270.00 mL solution of 0.00245 M CD5. What is pQsp for AD4? pQsp = -log(Qsp)
- A 110.00 mL solution of 0.00195 M A3B2 is added to a 160.00 mL solution of 0.00155 M C3D4. What is pQsp for A3D2?If a 25.00 ml aliquot of a solution containing sulfide ions was added to 25.00 ml of 0.0134 M Cu2+ solution. The CuS precipitate was washed and collected. The pH of the combined filtrate and washings was then adjusted by the addition of ammonia and the solution was titrated with 0.01453 M EDTA, using a suitable indicator. The endpoint volume was found to be 11.13ml. Find the molarity of the sulfide in the sample solution.1_ In the titration why the colour changes from wine red to bue? 2_what are the harmful effects of hard water? 3_write the principle involved in hardness of water by EDTA method?
- A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: Fe3++Cu+⟶Fe2++Cu2+Fe3++Cu+⟶Fe2++Cu2+ Complete the two half‑reactions that occur at the PtPt indicator electrode. Write the half‑reactions as reductions. half‑reaction: ?∘=0.161 V half‑reaction: ?∘=0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. ?E of the Ag∣∣AgClAg|AgClelectrode is 0.197 V.0.197 V. ?=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V E=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V ?=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V E=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V ?=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V E=0.767…Two hypothetical salts, LM2 and LQ, have the same molar solubility in H2O. If Ksp for LM2 is 3.20 × 10–5, what is the Ksp value for LQ?Two hypothetical salts, LM2 and LQ, have the same molar solubility in H2O. If Ksp for LM2 is 3.20 10–5, what is the Ksp value for LQ?
- At pH 7, what metal ions can be titrated by EDTA?Quantitative analysis of metal ions in pharmaceutical and cosmetic products can be done using direct and indirect complexometric titrations employing concepts of complex formation, masking, and blocking. For example, all Fe3+, Mg2+, Al3+ and Zn2+ can form a complex with EDTA at pH 10. However, at pH 5, Mg-EDTA complex formation is inhibited, whereas at pH 2, only Fe-EDTA complex formation is favored. Knowing these, an analyst tried to determine the % composition of a 0.1000 g sample containing soluble salts of Fe3+, Mg2+, and Al3+. The sample was initially dissolved in 250.0 mL of distilled water. To determine the total ion content of the sample, a 50.00 mL aliquot was buffered to pH 10.00 before adding 50.00 mL of 0.0500 M EDTA, and the excess EDTA was back titrated with 24.18 mL of 0.0750 M standard Zn2+ solution until the EBT endpoint. Another 50.00 mL aliquot was buffered to pH 2.00 and was added with small amount of SCN- producing blood-red Fe(SCN)2+ complex. This was titrated…A 0.1021 g sample containing ZnO was titrated using a standard EDTA solution with Erichrome Black T as an indicator. It took 25.52 mL of 0.0100 M EDTA to reach the endpoint. What is the percentage of ZnO in the sample?