Hi,
This problem was in my discussion section today and I'm confused about how to solve part b:
Find the pH of the following mixture:
a) .03 mols Ba(OH)2 in 1L water
b) 8.5 grams ammonia (NH3) added to the above solution (pKb ammonia 4.75)
Acid & Base Discussion Section Problem
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Re: Acid & Base Discussion Section Problem
I'm not sure how correct this is, but here is how I would approach the problem. From part (a), we know that Ba(OH)2 is a strong base, and 0.03 mols of it in 1L means that its concentration is 0.03M. Because there are 2 mols OH- per 1 mol of Ba(OH)2, you know the concentration of OH- will be 0.06M. For part (b), you first need to convert pKb to Kb with the calculation 10-4.75, which gives you a Kb of 1.78 x 10-5. From there, you set up the chemical equation, which looks like NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq). Then, you would convert the 8.5g of NH3 into moles of NH3, so because the molar mass of NH3 is 17g/mol, you would have 0.5M of NH3. Set up your ICE table with the initial concentrations of NH3, NH4+, and OH- being 0.5M, 0M, and 0M, respectively. You'd plug in the Kb value and eventually get that x = 0.003. Finally, you would add 0.003 to 0.06, take the negative log of that to get the pOH of the solution, and then subtract that result from 14 to get the pH. I believe the final answer should be a pH of 12.8
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Re: Acid & Base Discussion Section Problem
^ Josh's walkthrough looks correct. One thing I wanted to add is that, when considering what NH3 would do to the solution, we should know that NH3 is the conjugate base of a weak acid, NH4+. Because NH4+ is a weak acid, NH3 will somewhat take protons to form NH4+ and cause the OH- concentration to increase.
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