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Chapter 13: Problem 88

An aqueous solution has \(109.2 \mathrm{g} \mathrm{KOH} / \mathrm{L}\) solution. The solution density is \(1.09 \mathrm{g} / \mathrm{mL} .\) Your task is to use \(100.0 \mathrm{mL}\) of this solution to prepare \(0.250 \mathrm{m}\) KOH. What mass of which component, \(\mathrm{KOH}\) or \(\mathrm{H}_{2} \mathrm{O}\), would you add to the \(100.0 \mathrm{mL}\) of solution?

Short Answer

Expert verified
9.517 grams of water (H2O) need to be added to the 100 mL of solution to prepare 0.250 M KOH.

Step by step solution

01

Find mass of KOH in 100ml solution

1. First, convert the volume of the solution from ml to L by dividing by 1000, which gives \(0.1 \mathrm{L}\). Then, \nThe mass of KOH in the 100 ml solution can be calculated using the formula: \nMass = Volume (L) × Concentration (g/L) = \(0.1 \mathrm{L} × 109.2 \mathrm{g}/\mathrm{L} = 10.92 \mathrm{g} \)
02

Calculate the mass of KOH needed in desired solution

To find the mass of KOH required for a 0.250 M KOH solution in 100ml (or 0.1 L), we can use the formula: \nMass = Molarity (M) × Volume (L) × Molar Mass (g/mol) \nAs the molar mass of KOH is approximately 56.11 g/mol, we find \nMass = \(0.250\, \mathrm{M}\, ×\, 0.1\, \mathrm{L}\, ×\, 56.11\, \mathrm{g/mol}\) = 1.403 \mathrm{g} \nThis is the mass of KOH we need in our 0.250 M KOH solution.
03

Find the mass of the component to be added

Subtract the mass of KOH in desired solution found in step 2 from mass of KOH in the initial solution found in step 1. So, the mass difference is \(10.92 \mathrm{g} - 1.403 \mathrm{g} = 9.517 \mathrm{g}\). Since we get a positive number, this means that we have excess KOH in our solution. As a result, we would need to add water (H2O), not KOH to our solution to achieve the desired molarity.

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