Question

A solution contains \(158.2 \mathrm{~g}\) of KOH per liter; its density is \(1.13 \mathrm{~g} / \mathrm{mL}\). A lab technician wants to prepare \(0.250 \mathrm{~m}\) KOH, starting with \(100.0 \mathrm{~mL}\) of this solution. How much water or solid KOH should be added to the 100.0 -mL portion?

Short answer

Answer: 1028 mL of water should be added to the 100.0 mL concentrated KOH solution to prepare a 0.250 M KOH solution.

Step-by-step solution

Determine the molarity of the concentrated KOH solution

First, let's find the molarity of the concentrated KOH solution. We are given that it contains 158.2 g of KOH per 1 L (1000 mL) of solution. The molar mass of KOH is (39.1 g/mol for K) + (15.999 g/mol for O) + (1.008 g/mol for H) = 56.107 g/mol. Molarity (M) of KOH in the concentrated solution can be calculated by: M = (mass of solute in grams) / (molar mass of solute * volume of solution in liters) M = (158.2 g) / (56.107 g/mol * 1 L) M = 2.82 M (approximately) So, the concentrated KOH solution has a molarity of 2.82 M.

Calculate the moles of KOH in the initial 100 mL solution

Next, we need to find out how many moles of KOH are present in the given 100 mL of concentrated KOH solution. We will use the formula: moles of solute = molarity * volume of solution moles of KOH = 2.82 M * (100 mL / 1000) moles of KOH = 0.282 moles

Find the final volume of the diluted solution

Knowing the number of moles of KOH in the given solution, we can now determine the final volume of the solution with the target molarity of 0.250 M. Volume of solution = moles of solute / target molarity Volume of solution = (0.282 moles) / (0.250 M) Volume of solution = 1.128 L (1128 mL approximately)

Find the amount of water or solid KOH to be added

Now that we know the final volume of the diluted solution, we can find the amount of water or solid KOH needed to be added to the initial 100 mL of concentrated KOH solution. Amount of water to be added = final volume of solution - initial volume of solution Amount of water to be added = 1128 mL - 100 mL Amount of water to be added = 1028 mL The lab technician should add 1028 mL of water to the 100.0 mL concentrated KOH solution to prepare a 0.250 M KOH solution. No solid KOH should be added as the solution is already a mixture of KOH in water.