Need help with pH problem

[bjblenglish]

2KOH + H₂SO₄ > K₂SO₄ + 2H₂O (BALANCED PROBLEM)

In a typical acid/base chemical reaction, potassium hydroxide reacts with sulfuric acid to produce water, potassium sulfate and hydrogen gas. 150ml of 5.0 M potassium hydroxide is added to an ample amount of sulfuric acid. At the conclusion of the reaction, the total amount of water in the beaker is 418 ml.
?1. What is the molarity of potassium sulfate produced?
?2. What is the % yield if 21.3 grams of potassium sulfate was collected?
?3. Assuming KOH is a strong base. What is the pH of the potassium hydroxide used in the solution?
(I don't even know where to begin. I was told there is a formula for problem 3 on the internet any idea? This is for extra credit on our final today! Know one has answered the problem right yet. I would love to know how to do the problems) Thanks!!!

[Zedekiah]

To solve this problem with the information given, we have to assume the reaction has gone to completion entirely. Otherwise, we have no idea how much sulfuric acid was originally in the reaction mixture. It appears to me that the question is ambiguous.

1. I came up with 0.90 M potassium sulfate. 0.15 * 5.0 M = .75 moles of KOH. For every mole of KOH, you only have a half mole of K₂SO₄. This means you'll get .375 moles of K₂SO₄. Since you have .418 L and Molarity is equal to moles over liters, a simple division gives you 0.897, or 0.90 M with the right sig figs.
2. 21.3 grams needs to be converted to moles. The molar mass of potassium sulfate is 174.259. This gives a mole quantity of .122. This gives a yield of 33%. Dismal!
3. Here are some equations you'll be needing down the road :
14 = pOH + pH
pOH = -log [OH]
[OH] in this case is 5.0 M, so pOH = -log 5
pOH = -0.698970004!! Ouch, that's some basic stuff. Don't get any on your skin

14 + 0.698970004 = 14.7, the pH of your KOH solution.

Hope that helps!