Solution 1: 2 ml of KSCN Stock solution + 18.0 ml of Fe(no3)3 stock solution Sol

Question

Solution 1: 2 ml of KSCN Stock solution + 18.0 ml of Fe(no3)3 stock solution

Soln 2: 10ml of solution 1 + 10ml of Fe(no3)3 stock

Solution 3: 10ml of solution 2 + 10ml of Fe(no3)3 stock

Solution 4: 10ml of solution 3 + 10ml of Fe(no3)3 stock

Solution 5: 10ml of solution 4 + 10ml of Fe(no3)3 stock

Molarity of KSCN stock solution = 0.002M

Feno33 stock solution= 0.2M

I have to figure out the initial concentration of SCN- in each of the five solutions. This equation was given I think to help find it:

[scn-]ini= ([scn- stock]•Vstock)/Vtotal

Im specifically having trouble with finding the Vstock (volume of SCN stock within the solutions) in the equation. I figured out solution one but i cant figure out 2-5

Explanation / Answer

For Solution 1:

[SCN-] (solution 1) = [SCN-](stock) * V(stock)/Total volume

=> 0.0002M * 2/(2+18)

=> 2 * 10^(-5) M

For Solution 2:

[SCN-] (solution 2) = [SCN-](solution 1) * V(solution 1)/Total volume

=> 0.00002M * 10/(10+10)

=> 1 * 10^(-5) M

For Solution 3:

[SCN-] (solution 3) = [SCN-](solution 2) * V(solution 1)/Total volume

=> 0.00001M * 10/(10+10)

=> 5 * 10^(-6) M

For Solution 4:

[SCN-] (solution 4) = [SCN-](solution 3) * V(solution 1)/Total volume

=> 0.000005M * 10/(10+10)

=> 2.5 * 10^(-6) M

For Solution 5:

[SCN-] (solution 5) = [SCN-](solution 4) * V(solution 1)/Total volume

=> 0.0000025M * 10/(10+10)

=> 1.25 * 10^(-6) M

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