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Chapter 10: Problem 83

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

Short Answer

Expert verified
Answer: The technician should add 1.028 L (or 1028 mL) of water to the 100 mL KOH solution to achieve a concentration of 0.250 M KOH.

Step by step solution

01

Calculate the initial mass of KOH

Using the given mass concentration and volume, calculate the initial mass of KOH in the 100 mL portion. Mass concentration = 158.2 g/L Volume = 100.0 mL = 0.100 L (100 mL converted to liters) The initial mass of KOH = Mass concentration x Volume = 158.2 g/L x 0.100 L = 15.82 g Initially, we have 15.82 grams of KOH in 100 mL of the solution.
02

Calculate the number of moles of KOH

Have the molar mass of KOH equal 56.1 g/mol Calculate the number of moles of KOH using the mass of KOH and its molar mass: Moles of KOH = Mass of KOH / Molar Mass (KOH) = 15.82 g / 56.1 g/mol = 0.282 mol Now, we know that there are 0.282 moles of KOH in the original solution.
03

Determine the needed moles of KOH for final concentration

Now that we have to find how many moles would be needed to achieve the desired concentration of 0.250 M in the final solution: Desired concentration = 0.250 mol/L Original volume: 0.100 L Moles of KOH_required = Desired concentration x Original volume = 0.250 mol/L x 0.100 L = 0.025 mol To achieve the desired concentration of 0.250 M, we need 0.025 moles of KOH.
04

Determine if KOH needs to be added or water

Compare the needed moles of KOH for the desired concentration with the available moles of KOH in the original solution: Needed moles of KOH: 0.025 mol Available moles of KOH: 0.282 mol Since there are already more moles of KOH (0.282 mol) in the original solution than required for the desired concentration (0.025 mol), there's no need to add solid KOH. Instead, the technician should add water.
05

Calculate the final volume of the 0.250 M solution

Use the moles of KOH and the desired concentration to calculate the final volume of the solution: Final volume = Moles of KOH / Desired concentration = 0.282 mol / 0.250 mol/L = 1.128 L The final volume of the 0.250 M KOH solution is 1.128 L.
06

Calculate the amount of water required

Subtract the initial volume (100 mL) from the final volume (1.128 L) to find the amount of water to be added: Amount of water required = Final volume - Initial volume = 1.128 L - 0.100 L = 1.028 L The technician should add 1.028 L (or 1028 mL) of water to the 100 mL KOH solution to achieve a concentration of 0.250 M KOH.

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