The % w/w in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end p AgNO3 + Agl (s) + NO3 Ag+SCN-AgSCN What is the % w/w in the sample? O 0.9400 119.3 17.76 0.6860

The % w/w in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end p AgNO3 + Agl (s) + NO3 Ag+SCN-AgSCN What is the % w/w in the sample? O 0.9400 119.3 17.76 0.6860

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 79E: In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The...

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In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN- goes to completion, producing the Ag(CN)2- complex. Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of Ag(CN)2-. How many grams of NaCN were in the original sample?
The % w/w I– in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end point. AgNO3 + I– → AgI (s) + NO3– Ag+ + SCN– → AgSCN How many millimoles of AgNO3 reacted with I–? How many milligrams of I– (MM=126.9 g/mol) were present in the sample? What is the % w/w I– in the sample?
The % w/w I– in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach endpoint. AgNO3 + I– → AgI (s) + NO3– Ag+ + SCN– → AgSCN How many millimoles of AgNO3 reacted with I–?
Calculate the concentration of Fe2O3 in a mineral sample after solubilization of 0.4891g and availability of Fe as Fe2+. The released Fe2+ ions were titrated with a 0.02153 mol/L K2Cr2O7 solution requiring 36.92 mL to reach the end point of the titration. Show the calculations and express the concentration in % m/m. K2Cr2O7; + 6Fe2+ + 14H+ → 2Cr+3 + 2K+ + 6Fe3+ + 7H20
The %w/w Cl- (36.45 g/mol) in a 0.5785 g sample was determined by Volhard titration. After adding 65.00 mL of 0.0250 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.0365 M KSCN, requiring 15.00 mL to reach the end point. Report the %w/w Cl- in the sample
0.0585 g Na2C2O4 10 mL to adjust KMnO4 solution prepared in 0.1 M'pure water, 2 M H2SO4 added, heating process and 8.4 mL titrant as a result of titration it's spin out. Calculate the actual concentration of potassium permanganate accordingly
3 L contaminated air 50 mL 0.0116 M in an air pollution analysisCarbon dioxide (CO2) BaCO3 is passed through Ba (OH) 2 solution.is precipitated as. Excess of base, next to phenol phthalate (f.f.) indicatorIt is titrated with 23.6 mL of 0.0108 M HCl. CO2 in this air sampleCalculate its concentration in ppm. (Density of CO2Take it as 1.98 g / L. C = 12, O = 16 g / mol).
A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(II) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe2O3 (Molar mass= 159.69). 5Fe+2 + MnO4- +8H+ → 5Fe+3 + Mn+2 + 4H2O
A 0.3285 g sample containing chloride and inert material is titrated with 0.1012 M AgNO3, requiring 34.99 mL to reach the Ag,Cro. end point. Titration of a blank sample consumes 0.32 mL of the AgNO, titrant. Determine the % w/w chloride in the sample.
A 5.00 mL aqueous sample containing hydrogen peroxide was diluted to 25 mL and analyzed by titration with permanganate: 2MnO4- + 5H2O2 + 6H+ → 5O2 + 2Mn2+ + 8H2O The sample required 42.8 mL of 0.0175 M permanganate to reach the end point. What is the concentration of hydrogen peroxide in the original sample?
A 0.574-g sample of a chlorocarbon compound was analyzed by burning it in oxygen and collecting the evolved gases in a solution of NaOH. After neutralizing, the sample was treated with 26.21 mL of a 0.28 M AgNO3 solution. This precipitated the chloride (Cl-) out as AgCl and left an excess of AgNO3. The excess AgNO3 was titrated with 0.122 M KSCN and required 24.91 mL to reach the endpoint in a Volhard titration. Calculate the % w/w Cl– (35.45 g/mol) in the sample. Provide your answer to 2 places after the decimal point and without units. Reactions: Cl– + Ag+ → AgCl(s) Reaction 1 Ag+ + SCN– → AgSCN(s) Reaction 2
Suitable treatment of a 2.566 g soil sample converts all arsenic to AsO43-. 50.00 mL of 0.082 M Ag+ was added to form Ag3AsO4 precipitate. The excess Ag+ was back-titrated with 21.8 mL of 0.024 M SCN- to reach the equivalence point Ag+ + SCN- → AgSCN(s) Solve for the % As (MW = 74.92 g/mol) in the sample.