CHEM 107 Lecture Notes - Lecture 50: Magnesium Hydroxide, Ph, Sodium Hydroxide

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11 Aug 2017
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Concentration of h3o+ determines ph ph = -log[h3o+] If you know ph, you can determine [h3o+] You can also determine [oh-] and poh poh = -log [oh-] 10-poh = [oh-] ph + poh = 14. [h3o+] [oh-] = 1. 0 x 10-14 ph ranges. 1 ph unit corresponds to a factor of 10 difference in acidity. Since this is a strong acid, the 0. 050 m is telling us the h3o+ concentration ph = -log[h3o+] ph = 1. 3. Since [oh-] is known: poh = -log (0. 040) = 1. 4 ph + poh = 14 ph = 14 - 1. 4 = 12. 6. Mg(oh)2 (s) + 2hcl (aq) mgcl2 (aq) + 2h2o (l) Just like all stoichiometry problems, you need to get moles of whatever you are starting with and use mole ratios to get what you are looking for. 0. 10 g mg(oh)2 x (1 mol mg (oh)2 / 58. 33g mg(oh)2) x (2 mol hcl / 1 mol.

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