what maximum volume of 3 M solution can be prepared from 1 L each of 1 M KOH and 6 M KOH solutions:
1, without adding water?
2, by using water?
(1) Without adding water
We know that the Molarity equation is given by:
M1V1 + M2V2 = M3V3, where M1,M2,M3 are concentrations and V1,V2,V3 are their respective volumes.
For the first solution, M1 = 6 (6M KOH), M2 = 1 (1M KOH), V1 & V2 need to be determined.
For the resulting solution, we want M3 to be 3 (3M KOH) and V3 = V1 + V2, which needs to be maximized.
6V1 + 1V2 = 3(V1+V2)
6V1 + V2 = 3V1 + 3V2
3V1 = 2V2
or V1 = 0.67V2
If we want the resulting volume to be maximum, V2 should be maximum. The maximum available volume of V2 is 1 L.
Hence, V1 = 0.67 * 1 = 0.67 L
Therefore, the maximum volume of 3M KOH obtained will be V3 = V1 + V2 = 0.67 + 1 = 1.67L.
(2) By adding water
If we were allowed to add water, the maximum yield will be when we mix both the solutions (1M & 6M) and then add water, to dilute the solution to 3M. When we mix both, the resulting molarity will be:
MV = M1V1 + M2V2
M(2) = 6(1) + 1(1)
M = 3.5M
The solution will be 2L of 3.5M KOH. We have to dilute this, to bring it 3M
Again, applying the molarity equation, we have:
M1V1 + M2V2 = M3V3
If we add water, Molarity will be zero, hence M2 = 0
M1 = 3.5, V1 = 2L, M3 = 3 (since desired conc is 3M), V3 needs to be calculated
Therefore, 3.5 (2) = 3 (V3)
Hence, V3 = 7/3 = 2.33 L
Therefore, we get 2.33 L of 3M solution, by adding 0.33L of water to the mixture.
Hence, max volume of 3M KOH obtained is 2.33 L.