Determining the Molar Solubility of Iron (II) Hydroxide

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Given QuestionCalculate the molar solubility of Fe(OH)2 (Ksp=4.87*10^-7). 

Tutor’s Note: The Ksp radical that’s written down comes in later, I got confuzzled by my own notes! Happens to all of us. Sorry students 🙂 

Step 1: Write out the reaction equation with all ions. Make sure it’s balanced! We’ll use this in the next step with our ICE table.

Step 2: Create an ICE table for the reaction. For every mole of Fe(OH)2 “lost” in the reaction, we gain 1 mol Fe2+ and 2 moles OH. This will be our algebraic part of the ice table, where we lost -s amount of Fe(OH)2, and gained +s of Fe ions and +2s of OH molecules.

Step 3: We can  now calculate our Ksp value! We know that Ksp=[products] or the concentration of our products. Without calculating experimental values, we can use our algebraic interpretation of the equation to solve it. Since we know that [Fe2+]=s and [OH-]=2s, we can set up the equation so that Ksp=[Fe2+]*[OH-]^2. Insert our expressions for these and we get that the ksp= 4s^3. We get the four because we square the OH ion.

Step 4: We can now set this to find s, or the molar solubility. This is where my mistake comes back in! I thought it was a given equation, but it’s what we calculate when we need to gets ‘s’. Your final answer should be 2.30*10^-6 M.

As always if you have questions, comments, or concerns, don’t be afraid to contact us! Check my numbers, my procedure, and let me know if there are any mistakes.

Thanks everyone! Happy studies.

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