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1 MITOCW MIT3_091SCF10Final_Exam_A_Prob_10_300k SPEAKER: The following content is provided under a Creative Commons License. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JOCELYN: Hi. Jocelyn here. And today we're going to go over fall 2009's final exam, problem 10. As always we're going to start by reading the question. You have 333 milliliters of alkaline solution at a ph equal to 9.9. You wish to neutralize this by reacting it with 222 milliliters of acid. What must be the value of the ph of the acid? So first off let's write down the information that was given and the information that he's asking for. So we have 333 milliliters of a solution with ph equal to 9.9. And he's asking you what must the ph be if you want to put in 222 milliliters in order to completely neutralize that basic solution. And the key here is to recognize what neutralize means. And a neutralization is when you take acid, or protons, and base and it reacts to form water, thus creating a neutral solution. So the next step would probably be to figure out how much excess hydroxide we have in this solution. And then we can figure out how much hydrogen we need to react with that hydroxide and to form water in order to completely neutralize this solution. So first step is to find out what our concentration of OH minus is. And we do that by remembering a few things about acid/base solutions. Right? We know that if you have the ph plus the poh it equals 14. And this is from the equilibrium constant of this neutralization reaction. And then we can figure out what the poh is because we're given the ph of our initial solution and that equals Now using the definition of poh, which is that the poh equals the negative log of the concentration of the OH minus, we can rearrange that and get that our OH minus concentration equals 10 to the negative poh. Right? Just using our log rules there. So this gives us that our current concentration of the hydroxide ion is 10 to the negative 4.1. That gives us the concentration of the hydroxide, but what we really want to know, if we're going to use our neutralization reaction here, is we want to know how many moles of hydroxide we have in that solution. And we can just do that with stoichiometry. Right? The concentration is in moles per liter. And we know-- so our moles hydroxide equals our concentration, which is 10 to the negative 4.1 moles per liter. Times-- if we want to get rid of that liter unit we multiply it by our volume, which is 333 milliliters, converting it from milliliters to liters. And this gives us 2.65 times 10 to the negative 5 moles of OH minus in solution.

5 ionization of ammonia. OK. Moving to part iii. We have ammonium chloride, and this will dissociate into ammonium and chloride. And even if you're not sure how much it will dissociate, it will still dissociate to some amount. So it will have at least a little bit of an effect on our initial ionization reaction. So here we see the dissociation produces ammonium ions, and just like the first case that is one of the products in our ionization reaction. And so Le Chatelier's Principle, common ion effect, whatever you want to call it, tells us that it will shift the equilibrium to the left and decrease ionization. So with that we're done with problem 10 from the final. It had a lot to do with acid/bases, so hopefully you feel more comfortable with them after this.

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