The % (w/w) I- (MM = 126.9) in a 0.5812-g sample was determined by Volhard titration. After adding 50.00 mL of 0.06912 M AgNO3 and allowing the precipitate to form, the remaining silver was back-titrated with 0.07012 M KSCN, requiring 36.17 mL to reach the endpoint. Report the %(w/w) I- in the sample. 17.76 % 20.08 % 35.52 % 40.16%

The % (w/w) I- (MM = 126.9) in a 0.5812-g sample was determined by Volhard titration. After adding 50.00 mL of 0.06912 M AgNO3 and allowing the precipitate to form, the remaining silver was back-titrated with 0.07012 M KSCN, requiring 36.17 mL to reach the endpoint. Report the %(w/w) I- in the sample. 17.76 % 20.08 % 35.52 % 40.16%

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.35QAP

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The % (w/w) I- (MM = 126.9) in a 0.5812-g sample was determined by Volhard titration. After adding 50.00 mL of 0.06912 M AgNO3 and allowing the precipitate to form, the remaining silver was back-titrated with 0.07012 M KSCN, requiring 36.17 mL to reach the endpoint. Report the %(w/w) I- in the sample. 35.52 % 20.08 % 17.76 % 40.16%
A mineral in a fine state of division (0.6324 g) was dissolved in 25.0 mL of 4.0 mol / L boiling HCl and diluted with 175.0 mL of H2O containing two drops of methyl red indicator. The solution was heated to 100 ° C and a heated solution containing 2.00 g of (NH4) 2C2O4 was added slowly to precipitate CaC2O4. Next, NH3 6.0 mol / L was added until the indicator changed from red to yellow, indicating that the liquid was neutral or slightly basic. After slow cooling for 1 hour, the liquid was decanted, the solid transferred to a crucible and washed five times with 0.10 wt% (NH4) 2C2O4 solution, until no Cl- was detected in the filtrate with the addition of AgNO3 solution. The crucible was dried at 105 ° C for 1 hour and then taken to an oven at 500 ° C ± 25 ° C for two hours. The mass of the empty crucible was 18.2311 g. The crucible mass with CaCO3 (s) was weighed 5 times to an average of 18.5467 g. Determine the percentage, by mass, of Ca in the mineral. Ca (40.078 g / mol); C (12.01078…
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–?
A 0.1214-g sample of impure Na2CO3 was analyzed by the Volhard method. After adding 50.00 mL of 0.07011 M AgNO3, the sample was back titrated with 0.06021 M KSCN, requiring 27.15 mL to reach the end point. State the net ionic equation.
10mL of a 10% by weight MgCl2 solution (density = 1.1 g / mL) is precipitated as magnesium ammonium phosphate after necessary processes, filtered and washed. The precipitate is dissolved in 50mL 1M HCl and excess acid is titrated with 2.0M NaOH solution in the presence of methyl orange. Find the NaOH consumption (Mg = 24,3g / mol, Cl = 35,5g / mol)
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A sample is analyzed for chloride by the Volhard method. From the following data, calculate the percentage of chloride present:Weight of sample = 6.0000 g dissolved and diluted to 200 mLAliquot used = 25.00 mL AgNO3 added = 40.00ml of 0.1234MKSCN for back titration = 13.20ml of 0.0930M
1ml of 1mg/ml of diluted aspirin powder solution is titrated by 0.005M NaOH. Before titration, 5 ml ethanol, 14 ml CO2 free water, and 4 drops bromothymol blue indicator is added into diluted aspirin powder solution. Mass of 0.005M NaOH used is obtained by weight titration = 1.1384g . [Molar mass of aspirin =180.15]; for solution, 1g=1mL at 25 degree Celcius. Calculate the purity of the powder.
(a) Will Ca(OH)2 precipitate from solution if the pH ofa 0.050 M solution of CaCl2 is adjusted to 8.0? (b) WillAg2SO4 precipitate when 100 mL of 0.050 M AgNO3 ismixed with 10 mL of 5.0 x 10-2 M Na2SO4 solution?
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 the endpoint. Report the %(w/w) I- in the sample. [Ans. 17.76 % (w /w)]
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
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?