7. Draw a tiration curve for the analysis of 50.00mL solution of KOH, that is 0.1000 M and that for hydrazine is 0.0750 M. Calculate the pH after addition of 0.00, 20.00, 24.00, 25.00, 26.00, 44.00 and 50.00 mL of 0.2000 M HCIO4.
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- Consider the titration of 25.00 mL of 0.03555 M Co2+ by 0.02784 M EDTA at pH 10.00. Kf is 1045. Calculate the pCo2+ when 35.00 mL of EDTA are added.Chromel is an alloy composed of nickel, iron and chromium.A 0.6472 g sample was dissolved and diluted to 250 mL. When a50 mL aliquot of 0.05182 M EDTA was mixed with an equal volumeof the diluted sample and all the three ions were chelated, a 5.11 mLback titration with 0.06241 M copper (II) was required.The chromium in a second 50 mL aliquot was masked through theaddition of hexamethylenetetramine, titration of the Fe and Nirequired 36.28 mL of 0.05182 M EDTA.Iron and chromium were masked with pyrophosphate in a third50 mL aliquot and the nickel was titrated with 25.91 mL of theEDTA solution.Calculate the percentage of nickel, chromium and iron in thealloy.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?
- Generate a curve for the titration of 50.00 mL of a solution in which the analytical concentration of NaOH is 0.1000 M and that for hydrazine is 0.0800 M. Calculate the pH after addition of 0.00, 10.00, 25.00, 35.00, 45.00 and 50.00 mL of 0.2000 M HClO4.Calculate the alkaline strength of pearl ash ( impure potassium carbonate) in terms of percent K2O from the following data: Sample=0.3500 g; HCl used=48.03 m; NaOH used for back titration=2.02 mL; 1.000 mL HCl = 0.005300 g Na2CO3; 1.000 mL NaOH = 0.02192 g KHC2O4·H2ODetermine the E in each titrant volumes: 10.00 mL; 20.00 mL; and 30.00 mL, for the titrationof 20.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in a matrix of 1 M HClO4.In 1 M HClO4:Eo (Fe3+/Fe2+) = +0.767 VEo (Ce4+/Ce3+) = +1.70 V
- Calculate the alkaline strength of pearl ash (impure potassium carbonate) in terms of percent K2O from the following data: Sample = 0.3500 g; HCl used = 48.03 mL; NaOH used for back titration = 2.02 mL; 1.000 mL HCl = 0.005300 g Na2CO3; 1.000 mL NaOH = 0.02192 g KHC2O4.H2O.Compare the two quantities based on the given condition. Titration of the same sample solution of MgCl2 I. Volume at EP with 0.10 M EDTA as titrant II. Volume at EP with 0.10 M AgNO3 as titrant Are they equal? Or which one is greater? Or can it be determined?after dissolving in 50.0 ml of good water, the zinc in a 0.3232 gram sample of foot powder was titrated at a pH=11.0 with 19.67 ml of a 0.01222M EDTA solution. calculate the %Zn in this sample
- Consider the titration of 100.0 mL of 0.010 0 M Ce4+ in 1 M HClO4 by 0.040 0 M Cu+ to give Ce3+ and Cu2+, using Pt and saturated Ag | AgCl electrodes to find the end point. (a) Write a balanced titration reaction. (b) Write two different half-reactions for the indicator electrode. (c) Write two different Nernst equations for the cell voltage. (d) Calculate E at the following volumes of Cu+: 1.00, 12.5, 24.5, 25.0, 25.5, 30.0, and 50.0 mL. Sketch the titration curve.(ii) Consider the titration of 25.00 mL of 0.02000 M CaSO4 with 0.01000 M EDTA at pH 10.00. Write the chemical equation for this titration and calculate the conditional formation constant for this reaction. (iii) From (ii) calculate the concentration of Ca2+ and pCa2+ at the volume of 20.0 mL of EDTA added.50.00 ml of a 0.0500 M Ni2+ solution, whose pH is adjusted to pH=11 and the solution is additionally 0.100 M with respect to oxalate, is titrated with 0.100 M EDTA. Calculate the pNi after 10.0 mL, 25.0 mL, and 30.0 mL of titrant have been added.