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*
Answers: 3
= 0.0316 mol OH⁻Another question on Chemistry
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BONUS QUESTION
you are stuck with a problem. You need to measure pH of a solution known to be made...
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