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*

pH = 12.5

Explanation:

The reaction that takes place is

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:

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

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

Then we calculate the pOH of the solution:

And finally we calculate the pH:

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

→

i believe the answer is true.