Potassium permanganate is used to react with antimony (III) oxide, Sb Os, in a k
ID: 569360 • Letter: P
Question
Potassium permanganate is used to react with antimony (III) oxide, Sb Os, in a kinetics experiment. Mno,. (aq) 1.25 SbzO3 (aq) + 3H+ (aq) Mn2+ (aq) + 1.25 Sb205 (aq) + + 1.5 H2O (1) The reaction is very fast, so the temperature was lowered to a constant value of 5°C for the following experiments. The time needed for the purple color of MnO4 to disappear was recorded. Experiment mLs0.130MKMno, mL 0450 M SbOmLs total volume 5.00 10.00 5.00 Time (s) Rate 74 (8x 102 1.00 1.00 2.00 12.0 12.0 12.0 71 .ox lo2 6. Calculate [MnO4] under the initial conditions of experiment 1. 7. Calculate [Sb203] under the same conditions. 8. Calculate the average reaction rate, in M/s, for the disappearance of MnO4 during Experiment1. 9. Determine the rate law, with proper exponent orders, using the results of the 3 reactions. Note that in Experiment 3, the change in [MnO4] is twice as large as in the other experiments. You may leave the rate constant as the generic variable k for now. 10. Using any one of the experiments, calculate the value of the rate constant k.Explanation / Answer
NaOH reacts with HCl as per the reaction
NaOH (aq) + HCl (aq) --------> NaCl (aq) + H2O (l)
As per the stoichiometry of the reaction,
1 mole NaOH = 1 mole HCl.
12) Moles of HCl present initially = (volume of HCl taken in L)*(molarity of HCl) = (25.00 mL)*(1 L/1000 mL)*(0.1368 M) = 0.00342 mole (ans).
13) As per the stoichiometry of the reaction, NaOH reacts with HCl in a 1:1 molar ratio; hence, the mole(s) of NaOH dispensed by the buret = mole(s) of HCl present initially = 0.00342 mole (ans).
14) Volume of NaOH dispensed from the buret = (final volume) – (initial volume) = (34.31 mL) – (0.52 mL) = 33.79 mL = (33.79 mL)*(1 L/1000 mL) = 0.03379 L.
Molarity of NaOH = (moles of NaOH)/(volume of NaOH in L) = (0.0342 mole)/(0.03379 L) = 1.012 mol/L = 1.012 M (ans).
15) When an experimenter overtitrates the end-point of the NaOH/HCl titration by adding excess NaOH, then the volume of NaOH dispensed from the buret is high. However, the mole(s) of NaOH neutralized is equal to the mole(s) of HCl present, which is constant. Therefore, the calculated molarity = (moles of NaOH or moles of HCl)/(volume of NaOH dispensed in L) will be low.
16) The buret used to dispense NaOH in the titration experiment must be rinsed several times with the NaOH solution. This is done to make sure that there are no traces of moisture in the buret. Moisture of water will dilute the NaOH; therefore, a higher volume of NaOH will be required to reach the end-point.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.