Determine how you would prepare 220 mL of a 0.500 M potassium hydroxide solution starting with: a) solid potassium hydroxide b) a 1.25 M potassium hydroxide solution

a) 6.17 g of solid KOH

b) 88.0 mL of a 1.25 M potassium hydroxide solution

Explanation:

a) We have to determine the mass of solid potassium hydroxide that has to be weighed to prepare 220 mL of a 0.500 M potassium hydroxide solution. We will also require the molar mass of potassium hydroxide, KOH, 56.11 g/mol.

0.220 L × (0.500 mol/L) × (56.11 g/mol) = 6.17 g

b) We will use the dilution rule to determine the volume of the 1.25 M potassium hydroxide solution to be measured.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.500 M × 220 mL / 1.25 M = 88.0 mL

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explanation:

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