If 7 mL of water is added to 3 mL of a 0.50 M KOH solution, the concentration of KOH is changed to?
I am not sure how to go about doing this. Please help, thank you!!!
Would I do (7)(3) / 0.50?
Assuming the volume will add to give a final volume of 10 mL (technically they will not), then the new M is
0.50 x (3/10) = ?
Oh ok thank you!
To find the new concentration of KOH, you need to consider the total volume of the solution after adding water. In this case, you have 7 mL of water added to 3 mL of 0.50 M KOH solution.
To determine the new concentration, you'll first need to calculate the total volume of the solution. The water and KOH solution are mixed together, so the total volume of the solution is 7 mL (water) + 3 mL (KOH solution) = 10 mL.
Next, you can use the concept of molarity (M) to determine the concentration of the KOH.
The formula for molarity is:
Molarity (M) = Moles of solute / Volume of solution (in liters)
First, let's convert the volume into liters:
10 mL = 10 mL * (1 L / 1000 mL) = 0.01 L
To find the moles of KOH, you can use the formula:
Moles of solute = Molarity * Volume of solution (in liters)
Moles of KOH = 0.50 M * 0.01 L = 0.005 moles
Finally, to find the concentration of KOH after adding the water, you can divide the moles of KOH by the total volume of the solution:
Concentration = Moles of KOH / Total volume of solution (in liters)
Concentration = 0.005 moles / 0.01 L = 0.50 M
Therefore, the concentration of KOH remains the same at 0.50 M after adding 7 mL of water to 3 mL of the original KOH solution.