# Enthalpy, heat of neutralization, calorimetry

1. Jun 29, 2010

### maxbashi

1. The problem statement, all variables and given/known data
A quantity of 2.00x10^2 mL of .862 M HCl is mixed with 2.00x10^2 mL of .431 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.48 C.

For the process
H+(aq) + OH-(aq) --> H2O (l)
the heat of neutralization is -56.2 kJ/mol. What is the final temperature of the mixed solution?

2. Relevant equations
I'm thinking all I need is q=ms delta t

3. The attempt at a solution
q=(heat of neutralization per mol)(number of moles H2O)
q= -56.2 kJ/mol x (200 mL*.862mol/1000mL) = -9.69 kJ

If -9.69 = q = ms(T(f)-T(i))
Then T(f) = q/ms + T(i)
= -9690 J/(400g)(4.184g/J*C) + 20.48*C
= 121.8 degrees C

The answer should be 26.3 degrees C. Any help?

2. Jun 29, 2010

### Staff: Mentor

Do you know how to use calculator?

$$\frac{9690}{400\times4.184} + 20.48$$

and not

$$\frac{9690}{400} 4.184 + 20.48$$

(not to mention fact that you have lost minus sign, in a way luckily for you).

Last edited by a moderator: Aug 13, 2013
3. Jun 29, 2010

### maxbashi

well that's embarrassing. thanks.

4. Jun 30, 2010

Happens