1, when 10.0 mL of 0.020 M IO_3, 30.0 mL of starch indicator, 10.0 ml of 0.0080
ID: 981521 • Letter: 1
Question
1, when 10.0 mL of 0.020 M IO_3, 30.0 mL of starch indicator, 10.0 ml of 0.0080 M HSO_3; and 50.0 are mixed, what is the molar concentration of HSO_3 in the resulting 100.0 mL of solution (before any occurs)? See the experiment on QUANTITATIVE SOLUTION PREPARATION earlier in the lab manual Chapter 4 (Section 4) in the lecture text, if you have forgotten how to do dilution calculations. The multiple trials of this reaction make it possible to write a rate equation that describes this reaction: Rate = k [IO_3 ]^3 [HSO_3 ]^nExplanation / Answer
For Trial 1,
initial volume of [IO3-] = 10 ml of 0.02 M
moles of IO3- = 0.02 M x 10 ml = 0.2 mmol
final volume = 100 ml
final concentration of [IO3-] in 100 ml solution = 0.2 mmol/100 ml = 0.002 M
Similarly, for HSO3-,
initial volume HSO3- = 10 ml of 0.008 M
moles of HSO3- = 0.008 x 10 = 0.08 mmol
concentration [HSO3-] in 100 ml = 0.08/100 = 0.0008 M
We can do the restof calculations the same way,
Table filled :
Trial V of IO3- V of starch V of HSO3- V of H2O Final [IO3-] M Final [HSO3-] M
1 10 30 10 50 0.002 0.0008
2 15 30 10 45 0.003 0.0008
3 20 30 10 40 0.004 0.0008
4 30 30 10 30 0.006 0.0008
5 10 30 15 45 0.002 0.0012
6 10 30 20 40 0.002 0.0016
7 10 30 30 30 0.002 0.0024
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