# Doubts regarding solubility product problem

gracy
As I am also attaching solution along with the problem there is no point in posting this thread in homework forums .
In the following question (solution is also there) https://s29.postimg.org/omlnt73lz/IMG_20170407_090748.jpg https://s18.postimg.org/hsuxm5uvt/IMG_20170407_090805.jpg why are we taking volume= 2 ? I know it's a mixture of two solutions of equal volumes but it's no where mentioned in the question that volume of each solution is 1 litre .
Another query regarding the Same problem is that why are we even bothered about the volume?we are directly given the concentrations of Ag and Cl (in the options. just look at the units , it's M that means molarity which is unit of concentration.) If it would have been number of moles then we would have required to divide it by volume to get concentration. Please Clear my doubts.

Mentor
1. It doesn't matter what the volume is - as long as you mix two equal volumes when trying to calculate concentration after the dilution you will get something like $\frac V {V+V}$ where the V cancels out leaving you with $\frac 1 2$. As V cancels out, we can safely ignore it.

2. They never assumed the volume to be 1 L, they just halved the concentration for the reason explained above.

3. But then, actually there is nothing wrong with assuming volume of 1L (or any other). All that is important is that you assume equal volumes and 1L is much easier to use in further calculations than - say - 2.7641 gallons.

• gracy
gracy
1. It doesn't matter what the volume is - as long as you mix two equal volumes when trying to calculate concentration after the dilution you will get something like VV+VVV+V\frac V {V+V} where the V cancels out leaving you with 1212\frac 1 2. As V cancels out, we can safely ignore it.
If there will be mixture of three solutions of equal volume we will divide it by 3, right?

Another query regarding the Same problem is that why are we even bothered about the volume?we are directly given the concentrations of Ag and Cl (in the options. just look at the units , it's M that means molarity which is unit of concentration.) If it would have been number of moles then we would have required to divide it by volume to get concentration. Please Clear my doubts.