Calculating the ph of an unknown solution by measuring the precipitate 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.
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:
$$\ce{2 OH- (aq) + Pb(NO3)2 (aq) -&gt; Pb(OH)2 (s) + 2 NO3 (aq)}$$

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 $\left(K_\mathrm{sp} = 1.42 \times 10^{-20}\right)$.