I was wondering if there is some kind of equation out there to solve a problem like this. How many milliliters of a 25% solution of potassium chloride must be added to 8 milliliters of pure potassium chloride to obtain a 35% solution.

 

 

 

thanks

Can't be solved. The reason for this is that potassium chloride is an ionic solid. It is not a liquid so you can not have a pure solution of it. It would be like asking how many milliliters of Frosted Flakes do you need to add to a bowl of milk to get five hundred grams of cereal.

OK but mathematically how do you solve it, if it were possible?

youd find out the density of kcl when its moleten then figure out the mols and go from there i think

I'm pretty sure there is either an algebraic equation you can set up for a problem like this or a chemistry equation that you can use to solve it.

molten KCl.... HA!

 

If you replace KCl with something like say for example alcohol, think about it like this:

 

You have 8ml of 100% add V ml of 25% solution, and you end up with V+8 ml of 35% solution.

Of course this problem can be solved theoretically, but you have to make an important assumption.

The solid KCl must be completely compact (no air between the granules). Otherwise a statement about ml's of this solid is meaningless. Usually, with solids we talk about weight and your question then can be restated as follows:

 

"I was wondering if there is some kind of equation out there to solve a problem like this. How many milliliters of a 25% solution of potassium chloride must be added to ... grams of pure potassium chloride to obtain a 35% solution."

 

You need to know the density of a 25% solution of potassium chloride. You can measure this, or use some published table.

 

With that info the problem is simple. Suppose you have X grams of KCl, then you proceed as follows:

 

Suppose you have Y grams of a 25% solution of KCl and you add X grams to this and all of X dissolves (assuming that KCl is so soluble that a 35% solution can be made), then you have a liquid with (0.25*Y + X) grams of solution. The total weight of this is Y + X grams. So, you need to solve the following equation:

 

0.25*Y + X = 0.35*(Y+X)

 

Solving this is piece of cake, for a given Y.

You may use bulk density for loose KCl. Engineers do such things every day

 

Best,

Borek

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