Question

To initiate a reaction 8.00mL of 0.35M H2O2 is added to a beaker that contains 9.00mL...

To initiate a reaction 8.00mL of 0.35M H2O2 is added to a beaker that contains 9.00mL of a 0.350M KI solution. What is the initial concentration of H2O2 and KI in the beaker? M(H2O2)=______ M(KI)=_________

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Answer #1

In order to find initial concentration we use moles of each H2O2 and KI.

To find total volume we add both volumes.

We know concentration = mol solute /volume of solution in L .

Lets calculate

Moles of H2O2

Lets rearrange molarity equation to get moles of H2O2 from its molarity and volume.

n H2O2 = Volume of H2O2 solution in L x molarity of H2O2

= 0.008 L x 0.35 mol / L

= 0.0028 mol

n KI = 0.009 L x 0.350 mol/L

                = 0.00315 mol KI

Calculation of H2O2

[H2O2]= mol H2O2 / total volume in beaker

= 0.0028 mol / ( 0.008+0.009) L

= 0.165 M

[KI]= 0.00315 mol / ( 0.008+0.009) L

= 0.185 M

So the concentration of H2O2 and KI in the solution present initially are 0.165 M and 0.185 M respectively.

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