[swp]

The following is a problem that is giving me a hard time.  If anyone could please help me I would be greatly appreciated.

8. If you added 4.62g of solid potassium hydroxide to a 50.0mL solution of 1.6M sulfuric acid.
                a. what will the concentration of all ions be in the solution when the reaction is complete? Assume the final volume is 50.0mL.
                b. if there are only 7.40g of SO₄^2- ions produced, what is the %yield of this reaction?

[sdekivit]

the first and most important question to begin solving this and many more problems: what is the reaction equation with proper stoichiometry ?

[tamim83]

Well the first thing that you need is a balanced equation for your reaction. 

2KOH + H₂SO₄ ==> K₂SO₄ + 2H₂O

Now you want to write a total ionic equation.  Everything here but water is aqueous.  Also KOH is a strong base and sulfuric acid is a strong acid, so you can assume they completely ionize in solution

The next thing you need is to convert everything to moles.  For the sulfuric acid you can use M*V=Moles.    You can convert grams of potassium hydroxide to moles of potassium hydroxide using the molar mass. 

OK, you need to also find the limiting reagent so maybe find out how much potassium sulfate is produced with the moles of potassium hydroxide and sulfuric acid is produced.  The smallest amount is the winner, that will tell you how many moles of product you will get.  At the end of your reaction the LR will be completely consumed, and you are left with product and excess reagent.  You can get the moles of excess reagent ions by figuring out how much is consumed and subtracting it from the initial amount.  To get concentration, just divide everything by 0.05L. 

Hope this guides you a bit.  Good luck