What volume of a 2.70 M KI stock solution would you use to make 0.445 L of a 1.25 M KI solution?

Question

What volume of a 2.70 M KI stock solution would you use to make 0.445 L of a 1.25 M KI solution?

Answer 1

Notice that when you dilute a concentrated chemical solution with water, the number of moles of the solute remains the same. Since M=mol/L, M1*L1=M2*L2.

Answer 2

Hi Kendall, this problem can be solved using the formula M₁V₁ = M₂V₂.

From our given information, we know that:

M₁ = 2.70 M (concentration of the 1st stock solution, in Molarity (moles solute per liter solution))

V₁ = ? L (volume of 1st stock solution (units = Liters); this is our unknown, so we'll be solving for it!)

M₂ = 1.25 M (concentration of second stock solution, in Molarity)

V₂ = 0.445 L (volume of 2nd stock solution (units = Liters))

To solve, we just need to plug our know values into our formula, and solve for V₁:

M₁V₁ = M₂V₂

(2.70 M)(V₁) = (1.25M)(0.445L) divide both sides by 2.70M to isolate V₁:

V₁ = [(1.25M)(0.445L)] / (2.70) plug into calculator - solve for V₁:

V₁ = 0.206 L

Hope that helps!