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BONUS QUESTION

you are stuck with a problem. You need to measure pH of a solution known to be made from a metal hydroxide, but you don't have a meter or any indicators. You do happen to have some lead(II) nitrate that is soluble, and you remember that lead (II) hydroxide is insoluble. You add some to 1 liter of your own unkown solution and a precipitate forms. You add more unttil the precipitate stops forming and then a bit more just in case. After you filter and dry the precipitate, you have 3.81 grams of it. What was the approximate pH of the original solution?

*This is just a bonus question but I'm very confused on it so any help would be appreciated*

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maacastrobr

Answer:

pH = 12.5

Explanation:

The reaction that takes place is

  • 2OH⁻(aq) + Pb⁺²(aq) → Pb(OH)₂(s)

The OH⁻ species come from the metal hydroxide of the solution, and Pb(OH)₂ is the precipitate.

Now we convert 3.81 grams of Pb(OH)₂ into moles, using its molar mass:

  • 3.81 g ÷ 241.12 g/mol = 0.0158 mol Pb(OH)₂

Now we convert 0.0158 Pb(OH)₂ moles into OH⁻ moles, using the stoichiometric coefficients of the reaction:

  • 0.0158 mol Pb(OH)₂ * [tex]\frac{2molOH^-}{1molPb(OH)_2}[/tex] = 0.0316 mol OH⁻

With the given concentration (1 L), we calculate [OH⁻]:

  • [OH⁻] = 0.0316 mol / 1 L = 0.0316 M

Then we calculate the pOH of the solution:

  • pOH = - log[OH⁻] = 1.5

And finally we calculate the pH:

  • pH = 14 - pOH
  • pH = 12.5

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