Water was added to 175mL of a KOH solution until the volume was 250 mL and the molarity was 0.315 M. What was the molarity of the initial concentrated solution?

0.45 M

Explanation:

This is a dillution problem, where we have a solution in a initial state with a initial concentration. Now, this solution of 175 mL KOH, was added water until it reach 250 mL. When we add water we are dilluting the solution and so, the concentration is lower than the initial.

So the volume was changed, and the concentration too, but the only thing that remains constant in both cases is the number of moles. This number was not altered, so, we can safely assume that the moles of the solution would be the same as the initial, and therefore, we can use the following expression:

n₁ = n₂ (1)

And we know that the moles:

n = M * V

so, if we replace this expression into the first one we have:

M₁V₁ = M₂V₂ (2)

From here, we can solve for M₁ and we have:

M₁ = M₂V₂ / V₁ (3)

So all we have to do is replace the given data into this expression and solve for M₁:

M₁ = 250 * 0.315 / 175

M₁ = 0.45 MAnd this would be the initial concentration of KOH

0.45 M

Explanation:

Now we need to make use of the dilution formula

C1V1 = C2V2

C1 = initial molarity of the solution which is the unknown

V1 = initial volume of the solution which is 175 ml

C2 = final molarity of the solution = 0.315 M

V2 = final volume of solution = 250 ml

From C1V1 = C2V2 we have;

C1 = C2V2/V1

C1 = 0.315 * 250 / 175

C1 = 0.45 M