[rleung]

Hey,

I have a quick question about the following problem:

The heat of neutralization of HCl(aq) by NaOH(aq) is -55.90 kJ/mol. If 1.25 M NaOH(aq) is added to 25 mL of 1.54 M HCl (both initially at 23 degrees C) in a coffee cup calorimeter, what is the temperature of the final solution after the HCl has been completely neutralized?

Assume that the heat capacity is negligible and that no heat is lost to the surroundings.  

I understand how to set up the equation to find the answer:

(-55.90 kJ/mol)(0.03125 mol HCl) = - (g of the solution)(4.184 x 10^-3 J/K*g)(change in T)

but I am confused as to how I obtain the grams of the solution.  Should I assume that the liters of the HCl + liters of the NaOH required to neutralize the acid will equal the liters of the solution present, and then assume that Na+ and Cl+ will be negligible by using the density of water to figure out the mass of solution?  Thanks so much

Ryan

[Borek]

Final mass is a sum of masses regardless of density changes. You may assume initial densities of 1g/mL or you may use 1.0526g/mL for 1.25M NaOH and 1.0253g/L for 1.54M HCl. Choice is yours

In case you will be asked where did you get these densities from - tables are built into CASC calculator.

[rleung]

Thanks.  For the density of HCl, do you mean 1.0253g/mL instead of g/L?

Thanks again

Ryan

[Borek]

Yeah, g/mL, typo.