A 50.0 mL sample containing Cd2 and Mn2 was treated with 40.4 mL of 0.0500 M...

A 50.0 mL sample containing Cd2 and Mn2 was treated with 40.4 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 19.2 mL of 0.0300 M Ca2 . The Cd2 was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 18.2 mL of 0.0300 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?

Expert Solution

queen_honey_blossom answered 2 years ago

A 50.0-mL solution containing Ni2+ and Zn2+ was treated with 25.0 mL of 0.0452 M EDTA...

A 50.0-mL solution containing Ni2+ and Zn2+ was treated with 25.0 mL of 0.0452 M EDTA to bind all the metal. The excess unreated EDTA required 12.4 mL of 0.0123 M Mg2+ for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another 29.2 mL of Mg2+ was required for reaction with the liberated EDTA. Calculate the molarity of Ni2+ and Zn2+ in the original solution

Consider the titration of 50.0 mL of 0.0500 M Cu+ with 0.1000 M Fe3+ to give...

Consider the titration of 50.0 mL of 0.0500 M Cu+ with 0.1000 M Fe3+ to give Cu2+ and Fe2+ using Pt and standard H+/H2 reference electrode to find the end point. The standard electrode potentials are E0Cu+/Cu2+=0.161 V and E0Fe2+/Fe3+=0.767 V. Calculate the actual electrode potential, E, at the following volumes of Cu2+ (i) 15, (ii) 25, and (iii) 26 mL

A 50.0 ml sample of 0.100 M Ba (OH)2 is treated with 0.100 M HNO3. Calculate...

A 50.0 ml sample of 0.100 M Ba (OH)2 is treated with 0.100 M HNO3. Calculate the ph after the addition of the following volumes of acid. a). 25 ml of HNO3 b). 50 ml of HNzo3 c). 75 ml of HNO3 d). 100 ml of HNO3 e). 125 ml of HNO3

A 50.0 mL sample of 0.0645 M AgNO3(aq) is added to 50.0 mL of 0.100 M...

A 50.0 mL sample of 0.0645 M AgNO3(aq) is added to 50.0 mL of 0.100 M NaIO3(aq). Calculate the [Ag+] at equilibrium in the resulting solution. The Ksp value for AgIO3(s) is 3.17 × 10-8. [Ag+] =____ mol/L

In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid...

In an acid base titration experiment, 50.0 ml of a 0.0500 m solution of acetic acid ( ka =7.5 x 10^-5) was titrated with a 0.0500 M solution of NaOH at 25 C. The system will acquire this pH after addition of 20.00 mL of the titrant: Answer is 4.581 Please show all work

Consider the titration of 50.0 mL of 0.0500 M C2H5NH2 (a weak base; Kb = 0.000640)...

Consider the titration of 50.0 mL of 0.0500 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HIO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 6.3 mL pH = (c) 12.5 mL pH = (d) 18.8 mL pH = (e) 25.0 mL pH = (f) 30.0 mL pH =

A 45.00 mL mixture containing 0.0616 M Mn2 and 0.0616 M Pb2 is titrated with 0.0829...

A 45.00 mL mixture containing 0.0616 M Mn2 and 0.0616 M Pb2 is titrated with 0.0829 M K2S resulting in the formation of the precipitates MnS and PbS. Calculate pS2– at the following points in the titration. The Ksp values for MnS and PbS are 4.6 × 10-14 and 8.9 × 10-29, respectively. 17.2 mL -- ? 45.5 mL -- ? The second equivalence point -- ? 73.9 mL -- ?

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium...

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium chloride (NaCl) and 0.05000 M potassium iodide (KI), what is the pAg of the solution after 15.00 mL of AgNO3 is added to the solution? Ksp, AgCl (s) = 1.82 x 10-10; Ksp, AgI(s) = 8.3*10-17. b. Same titration as in (a), what is the pAg of the solution after 25.00 mL of AgNO3 is added to the above solution? c. Same titration as...

A solution containing 45.00 ml of 0.0500 M metal ion buffered to pH = 10.00 was...

A solution containing 45.00 ml of 0.0500 M metal ion buffered to pH = 10.00 was titrated with 0.0400 M EDTA. Answer the following questions and enter your results with numerical value only. Calculate the equivalence volume, Ve, in milliliters. Calculate the concentration (M) of free metal ion at V = 1/2 Ve. Calculate the fraction (αY4-) of free EDTA in the form Y4-. Keep 2 significant figures. If the formation constant (Kf) is 1012.00. Calculate the value of the...

A solution containing 40.00 mL of 0.0500 M metal ion buffered to pH = 12.00 was...

A solution containing 40.00 mL of 0.0500 M metal ion buffered to pH = 12.00 was titrated with 0.0400 M EDTA. Calculate the fraction (αY4-) of free EDTA in the form Y4−. Keep 2 significant figures.

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