How do I solve this problem on a reaction between Ba (OH)2 and H2SO4?

Question

30 mL of 0.10M Ba (OH)2 (aq) is added to 30 mL of 0.10 M H2SO4 and the solution is stirred quickly. The rise is temperate T 1 is measured. The experiment is repeated using 90 mL of each solution and the rise in temperature T 2 is measured. How do I determine the the relationship between  T 1 and T 2?

Answer 1

I see you are listing a series of questions, so instead of doing them all for you, I'll guide you. Hope that's ok.

 

Ba(OH)₂ + H₂SO₄ ==> BaSO₄ + 2H₂O   balanced equation

The balanced equation shows you that Ba(OH)2 and H2SO4 react in a 1:1 mole ratio.  So since they are both 0.10 M solutions, they will also react in a 1:1 volume ratio, i.e. 30 ml + 30 ml, or 60 ml + 60 ml or 90 ml + 90 ml, etc.

 

Since 90 mls is 3x the original 30 mls, don't you think the heat generated, will be 3x that generated when 30 mls are used?

 

Now, The question is, will the temperature also change, and if so, by what amount?  Look at q = mC∆T.  In experiment #1 and #2 the value of C is constant, so no need to include it.  The q in exp. # will be about 3x the q in exp. #1, BUT, the m in exp. #2 is also 3 x that in exp. #1 (180 mls vs. 60 ml).  So, what do think will be the difference between T1 and T2?