[erinw88]

Before we turned in our homework, our professor pointed out that one of the answers in the back of the book was incorrect (we turn in our homework for completion and use the answers to make sure we are doing the problems correctly). Now I am studying for the test, and I keep getting a different answer than the one he gave us. In fact, I keep getting the answer in the back of the book. So I am wondering, is the book or my professor wrong? Normally I would think that the book is, but I would like to be sure.

Here is the question...

When a 9.55g sample of solid sodium hydroxide dissolves in 100.0g of water in a coffee-cup calorimeter, the temperature rises from 23.6°C to 47.4°C. Calculate (Delta)H (in kJ/mol NaOH) for the solution process

NaOH (s) -----> Na+(aq) + OH-(aq)

Assume that the specific heat of the solution is the same as that of pure water.


Our professor said that the answer should be -41.7 kJ/mol, and the answer in the back (and the answer that I keep getting) is -45.7kJ/mol. I realized that you get -41.7kJ/mol when you use 100.0g as the mass of the solution, and you get -45.7 when you use 109.55g as the mass of the solution. It didn't make sense to me to use just the mass of water and leave out the mass of sodium hydroxide when both make up the total mass of the solution. Am I missing something?

[Borek]

I would go for 109.55.

[Rhile3]

You have to specify if the reaction exothermic or endothermic.  The system which is reacting (The Na(OH)) has to be treated separately from the H₂O because it produces heat(exothermic) while the H₂O gains heat.

The other factor is that the ions of Na(OH) do not have the same Heat Capacity as water.

The  :delta:H cannot include all of the matter inside the calorimeter.  A calorimeter is Adiobatic, which means heat cannot transfer into or out of the system(ideally). 

 :delta:H for the entire system will equal 0 by the definition of Adiobatic.

Proof:
 (Delta)H_H2O   =(C_H20)(100.0 g)( (Delta)t)
 (Delta)H_Na(OH)=(C_Na(OH))(9.55 g)(-(Delta)t)

 (Delta)H_Solution=  (Delta)H_H2O+(Delta)H_Na(OH)=0

[jennatbee]

In my experience with undergraduate general chemistry, when a solid is dissolved in water and a heat transfer occurs, we assume the c of the solution is the same for that of water.  Usually, though, the problem specifies that. 
I agree that it should be -45.7 kJ/mol, if we assume the specific heat of the solution does not change.  In that case, you would add the masses for the q (or deltaH) of solution.