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4. A 0.5255 g sample containing an unknown mass of SnCl2 is dissolved in 50.00 mL...
The Fe2+ (55.845 g/mol) content of a 2.264 g steel sample dissolved in 50.00 mL of an acidic solution was determined by tiration with a standardized 0.120 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 44.82 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per gram of steel Mn2+5 Fe34 H20 + 5 Fe2+ MnO8 H concentration g Fe/g steel A 1.969 g...
• CIU TEXAS INSTRUM 2. A 50.00 ml sample of solution containing Fe?ions is titrated with a 0.0216 M KMnO4 solution. It required 20.62 ml of the KMnO4 solution to oxidize all the Fe? ions to Fe ions by the reaction: MnO4 (aq) + Fe? ) -------> Mn(ed) Felco (unbalanced) a) What was the concentration of Fel.ions in the sample solution? b) What volume of 0.0150 M K2Cr2O7 solution would it take to do the same titration? The reaction is:...
A sample of an unknown containing Fe2+ in the dissolved sample requires 23.15 mL of a 0.0020 M KMnO4 solution to reach the end point of the titration. 1)Calculate the moles of KMnO4 reacted. 2) Based on the moles you calculated , calculate how many grams of Fe2+ are in the unknown solution.
1. (4p) In an acidic aqueous solution, Fe2+ ions are oxidized to Fe3+ ions by MnO4": 5Fe2+(aq) + MnO4 (aq) + 8H(aq) → 5Fe3+ (aq) + Mn2(aq) + 4H2O(1) In part A of the experiment, suppose that 1.067 g of Fe(NH4)2(SO4)2.6H2O(s) are placed in a 250 ml Erlenmeyer flask to which 20 mL of water and 8 mL of 3 M H2SO4(aq) are added. The solution was titrated to the end point by adding 26.89 mL of KMnO4(aq) from the...
3. 0.7253 g sample containing an unknown weak acid HA was dissolved in 50 mL water and titrated against 0.1555 M NaOH, requiring 48.11 mL to of NaOH to reach the end-point. During the titration reaction, the pH of the solution is 3.77 when half of the HA is neutralized and the equivalence-point pH is 8.33. (a) State two ways to standardize the NaOH used in the titration. (b) Suggest and explain an indicator that can be used in the...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of the unknown was treated with 25.00 mL of 0.2404 M EDTA. The excess EDTA was then back titrated with 8.52 mL of 0.0694 M Zn2+ to reach the equivalence point. What was the concentration of Ni (in unit of M) in the 50.00 mL sample? Please keep your answer to three decimal places.
4. A 2.86 g sample containing both Fe and V was dissolved under certain conditions and diluted to 200.00 mL. Fe2 and VO ions. The titration of this solution required 22.64 mL of0.1000 M Ce" to reach end point. A second 50.00 mL aliquot was passed through a Jones reductor to forn ions. The titration of the second solution required 42.66 mL of 0.1000 M Ce solution to reach an end point. Calculate the percentage of Fe and V in...
A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The titration required 30.0 mL of base to reach the equivalence point, at which point the pH was 8.68. a) What is the molecular weight of the acid? b) What is the pKa of the acid?
A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.03 mL of 0.04947 M EDTA to reach the end point. A 50.00 mL sample of the unknown was then treated with F- to mask the A13+. To the 50.00 mL sample, 25.00 mL of 0.04947 M EDTA was added. The excess EDTA was then titrated with 0.02164 M Mn2+. A total of 25.3 mL was required to reach the methylthymol blue end point. Determine pA13+ and...