Question

In: Chemistry

A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+...

A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+ leading to the reaction CH2(COOH)2 + 6Ce4+ + 2H2O → HCOOH + 2CO2 + 6Ce3+ + 6H+ After standing for 10 minutes at 60°C, the solution was cooled and the x’ss Ce4+ was titrated with 0.1M Fe2+, requiring 14.4 ml to reach the ferroin end point. Calculate the M of the malonic in the sample.

Solutions

Expert Solution

queen_honey_blossom answered 2 years ago
Next > < Previous

Related Solutions

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03742 M EDTA solution. The solution is then back titrated with 0.02190 M Zn2 solution at a pH of 5. A volume of 20.48 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05893 M EDTA solution. The solution is then back titrated with 0.02306 M Zn2 solution at a pH of 5. A volume of 19.89 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04503 M EDTA solution. The solution is then back titrated with 0.02327 M Zn2 solution at a pH of 5. A volume of 22.80 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05832 M EDTA solution. The solution is then back titrated with 0.02380 M Zn2 solution at a pH of 5. A volume of 18.49 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 5.00-mL aliquot of a solution that contains 3.57 ppm Ni2+ is treated with an appropriate...
A 5.00-mL aliquot of a solution that contains 3.57 ppm Ni2+ is treated with an appropriate excess of 2,3-quinoxalinedithiol and diluted to 50.0 mL. The molar absorptivity of a Ni2+- 2,3-quinoxalinedithiol solution at 510 nm is 5520 L mol-1 cm-1. What is the absorbance of the above diluted Ni2+- 2,3-quinoxalinedithiol solution at 510 nm in a 2.00-cm cell?
A solution of 100.0 mL of 1.00 M malonic acid (H2A) was titrated with 1.00 M...
A solution of 100.0 mL of 1.00 M malonic acid (H2A) was titrated with 1.00 M NaOH. K1 = 1.49 x 10-2, K2 = 2.03 x 10-6. What is the pH after 50.00 mL of NaOH has been added?
A) If a 10.0 mL solution of "acid rain" is measured to have a pH of...
A) If a 10.0 mL solution of "acid rain" is measured to have a pH of 2.75, what is the molar concentration of strong acid HNO3 formed? Hint: pH=-log[H+] and 10^-pH=[H+] is in molarity units. B) The total acid concentration (nitric + nitrous acid) of the 10.0 mL solution of "acid rain" is determined by an acid-base titration method. You add 5.00 mL of the titrant sodium hydroxide with a concentration of 0.0114 M to reach the endpoint. i. What...
What is the pH of a solution that is 0.0924 M in malonic acid? What will...
What is the pH of a solution that is 0.0924 M in malonic acid? What will be the equilibrium concentration of each of the three malonic acid species? pKa1 = 2.85 pKa2 = 5.70 pH: _________________ [H2A]: ____________________ [HA-]: _________________ [A2-]: _________________
A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1408 M solution of...
A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1408 M solution of NaOH at 25C. a.What is the PH of the HF solution before titrant is added? b.How many milliliters of titrant are required to reach the equivalence point? c.What is the PH at 0.5ml before the equivalence point? d.What is the PH at the equivalence point? e.What is the PH at 0.5 ml after the equivalence point?
A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1281M solution of NaOH...
A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1281M solution of NaOH at 25C. a.What is the PH of the HF solution before titrant is added? b.How many milliliters of titrant are required to reach the equivalence point? c.What is the PH at 0.5ml before the equivalence point? d.What is the PH at the equivalence point? e.What is the PH at 0.5 ml after the equivalence point?
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT