9.0 mL of 0.1 M phosphate buffer was combined with 0.2 mL of 62 Unit/mL horseradish...
9.0 mL of 0.1 M phosphate buffer was combined with 0.2 mL of 62 Unit/mL horseradish peroxidase, 0.3 mL of each indicator, and varying amounts of 0.003% H2O2. If a total of 0.2 mL of H2O2 was added to the mixture, what is the final [H2O2] in units of moles H2O2/L solution?
Homework Answers
Total solution = 9.0 + 0.2 + 0.3 + 0.2 = 9.7 ml
H2O2 = 0.003 % = 0.003 g in 100 g water
Molar mass of H2O2 = 34 g/mol
Moles of H2O2 = 0.003/34 = 8.8*10-5 mol
100 ml of solution contains Moles of H2O2 = 8.8*10-5 mol
0.2 ml of solution contains Moles of H2O2 = (8.8*10-5 /100)*0.2 = 1.76*10-7 mol
1.76*10-7 mol of H2O2 is present in 9.7 ml of total solution.
So, 1 L of solution would contain moles of H2O2 = (1.76*10-7 /9.7)*1000 = 1.8*10-5 mol/L
Final H2O2 solution = 1.8*10-5 mol/L
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