(a) is correct and the reasoning is correct.
(b) Without balancing any of the equations, let me just sketch it out.
H2O2 + MnO4^- ==> Mn^+2 + O2
The calculation follows and I have numbered each equation.
1)moles MnO4^- = M x L
2)moles MnO4^- x factor = moles H2O2.
3)grams H2O2 = moles H2O2 x molar mass H2O2.
4)%H2O2 = (grams H2O2/weight sample)*100 = ??
If the endpoint is overshot, then mL MnO4^- is too high so moles in 1) is too high.
moles MnO4^- in 1) makes moles H2O2 in 2) too high.
moles H2O2 in 2) too high makes grams H2O2 in 3) too high and grams H2O2 in 3 (and4) makes %H2O2 too high. So your gut feeling is right but this shows you how to do these. The easy way on any of the too high, too low, the same problem(s) is to write the equation(s) needed to calculate the final answer, then go through each one as I've done. It's all very logical.
please, help me.. i think i know the answers, but i just want to make sure! i'd love your feed back :)
1) an air bubble is trapped in the tip of a buret before a titration. during the titration, the bubble is passed from the buret tip. will the reported volume of titrant used be too high,too low, or unchanged as a result?
--> i think it will be too high because you're including the volume of the bubble (gas) too.
2) in the determination of H2O2 by titration with permanganate, the end point was overshot (too much titrant to analyte)what effect would this have on the reported amount of H2O2? Explain.
--> this one i'm not actually sure how to go about this. i think that the reported amount of H2O2 would be greater than it actually is, but i'm not sure.
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you're so wonderful! you're right, i'll use the equations from now on. thank you for enlightening me!
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