You have an unknown metal hydroxide, but no meters or indicators. You have lead (II) nitrate, which is soluble in it, and lead (II) hydroxide that is insoluble in it. You take one liter of the unknown substance and add lead (II) nitrate, causing the formation of a precipitate. You add more lead (II) nitrate until the precipitate stops forming, and then add a little more. After you filter and dry the precipitate, you have 3.81 grams of it. What was the approximate ph of the original solution?

I know that the metal hydroxide would be a base, but I’m having a hard time figuring out anything else about this problem. My first thought was to figure out the molar mass of the precipitate to try and determine how much lead nitrate was added to the unknown substance, but I can’t figure that out if I don’t know what the precipitate is composed of. I thought about figuring out the molarity of the total solution to figure out how much lead nitrate was added, but again I couldn’t see a way to do it without more information.

I also thought, since we’re working with pH, we’re supposed to do something with dilution and changing powers, but I don’t see any way to do it.

This problem has me completely stuck. I just don’t see where to start. Any help would be greatly appreciated.

  • $\begingroup$ HINT - Write out the chemical reaction. $\endgroup$ – MaxW May 7 '19 at 16:49
  • $\begingroup$ @MaxW I tried it, but I’m still not seeing anything. I think I’m misunderstanding the reaction between the unknown substance and the lead nitrate. The only thing we’ve been talking about in class recently is Acid-base reactions forming salt and water or conjugate acids and bases, but i think there’s something I’m missing. $\endgroup$ – Screaming Taco May 7 '19 at 19:10
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    $\begingroup$ Does lead hydroxide precipitate in an acidic solution or a basic solution? // You need to figure out what the pH is of the solution, not what the unknown cation is. $\endgroup$ – MaxW May 7 '19 at 19:32
  • $\begingroup$ @MaxW That makes a lot of sense. Thank you very much for your help! $\endgroup$ – Screaming Taco May 7 '19 at 20:37
  • $\begingroup$ Could you add those equations to the question so we can see how far you got? $\endgroup$ – Karsten Theis May 10 '19 at 19:35

You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate $\mathrm{pOH}$ of the solution as $\mathrm{pOH} = -\log \ce{[OH-]}$. Then, $\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}$. Now, you see, since $\mathrm{p}K_\mathrm{w} = 14$, only unknown is $\ce{[OH-]}$.

Now look at the reaction of your unknown hydroxide with lead (II) nitrate: $$\ce{2 OH- (aq) + Pb(NO3)2 (aq) -> Pb(OH)2 (s) + 2 NO3 (aq)}$$

Since you know the precipitated amount of $\ce{Pb(OH)2}$ isolated from $\pu{1.0 L}$ of unknown hydroxide solution, you can calculate original $\ce{[OH-]}$:

$$\text{Amount of } \ce{OH-}\text{ in original solution} \\ = \pu {3.81 g}\text{ of } \ce{Pb(OH)2} \times \frac{\pu {1 mol}\text{ of } \ce{Pb(OH)2}}{\pu {241.21 g}\text{ of } \ce{Pb(OH)2}} \times \frac{\pu {2 mol}\text{ of } \ce{OH-}}{\pu {1 mol}\text{ of } \ce{Pb(OH)2}} \\ = \pu {0.0316 mol}\text{ of } \ce{OH-} $$ Thus, $ \ce{[OH-]} = \pu{0.0316 M}$, and hence, $$\mathrm{pH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}= 14 + \log(0.0316) = 12.5 $$

Asking is approximate $\mathrm{pH}$ of orignal solution. Thus, this answer is acceptable. However, keep in mind that trace of $\ce{OH-}$ will remain in solution due to the solubility of $\ce{Pb(OH)2}$ in water $\left(K_\mathrm{sp} = 1.42 \times 10^{-20}\right)$.

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