[karina]

I need to calculate the enthalpy of neutralization per mole of water produced...

50 ml of 1.0M HCl
50 ml of 1.0M NaOH
Initial temperature of both solutions= 28 degree celsius
Temperature after mixing = 35 deg. celsius

HCL + NaOH --> NaCl + H20

given information in the lab manual for this question: The density of the o.5M NaCl produced is 1.02 g/ml, and its specific heat is 4.04Jg-1K-1.

here's what i did:
amount of NaOH reacted= (0.05/1000)x 1.0 = 0.05mol
amount of HCl used = 0.05mol

0.05mol NaOH x (38.99g/1mol) = 1.95g NaOH
heat evolved = 4.18 x 1.95 x 7 = 57.057 J
Amount of OH- reacted = 0.05mol
amount of water formed =1:1ratio therefore, 0.05mol H20

so heat evolved per mole of water, heat of neutralization = 57.057 / 0.05 = 541.14 J


What's wrong here?..I didnt use the o.5M NaCl  and the 1.02g/ml and the 4.04Jg-1K-1 anywhere...I dont know what to do with it..

[karina]

The heat of solution  = 4.04 x 7 x 100 =2828J ?

I Know the heat gained by the calorimeter = 311.5 J

so heat of reaction = 2828 +311.5 =3139.5 J

Heat of neutralization per mole of water formed=3195.5/moles of water = 3195.5/0.05 = 62.79KJ/mol?

is this right?...i'm really confused.

[Rabn]

I don't have the time work through it at the moment but I want to ask if you remembered to take into account that the volume of the final mixture is 100mL.  What you are forgetting to take into consideration is how the sodium chloride in the solution effects the specific heat of the total solution.  You used the value for pure water in your calculations.  You need to take the salt into consideration. One thing about those folk that write the labs...they don't give you spurious data.  If they provide you with some data you are sure to need it.

[karina]

density = m/v
 m = d*v = 1.02 *100 = 102g
q = mc*delta t = 102*4.04*7 = 2884.5J

so heat ofneutralisation per mole of water = 2884.5 / 0.05 = 57.691 KJ/mol

but i still dint use the 0.5M NaCl anywhere...

[Rabn]

density = m/v
 m = d*v = 1.02 *100 = 102g
q = mc*delta t = 102*4.04*7 = 2884.5J

so heat ofneutralisation per mole of water = 2884.5 / 0.05 = 57.691 KJ/mol

but i still dint use the 0.5M NaCl anywhere...

By writing that are you saying that there are 102g of NaCl?  To find the specific heat of the solution you need to, i believe, use the mole fraction times the specific heat of the substance. i.e. if the NaCl has a mole fraction of .1 you would say (.1 * 4.04) + (.9 * (Cp water)) to get the specific heat of your solution.  THen calculate q etc...

[DrCMS]

but i still dint use the 0.5M NaCl anywhere...

Yes you did you used the weight and specific heat capacity of it here

density = m/v
 m = d*v = 1.02 *100 = 102g
q = mc*delta t = 102*4.04*7 = 2884.5J

[DrCMS]

density = m/v
 m = d*v = 1.02 *100 = 102g
q = mc*delta t = 102*4.04*7 = 2884.5J

so heat ofneutralisation per mole of water = 2884.5 / 0.05 = 57.691 KJ/mol

but i still dint use the 0.5M NaCl anywhere...

By writing that are you saying that there are 102g of NaCl?  To find the specific heat of the solution you need to, i believe, use the mole fraction times the specific heat of the substance. i.e. if the NaCl has a mole fraction of .1 you would say (.1 * 4.04) + (.9 * (Cp water)) to get the specific heat of your solution.  THen calculate q etc...

Rubbish the figure given it the specfic heat capacity of 0.5M NaCl solution, that all you need for the cacluation.

[karina]

density = m/v
 m = d*v = 1.02 *100 = 102g
q = mc*delta t = 102*4.04*7 = 2884.5J

so heat ofneutralisation per mole of water = 2884.5 / 0.05 = 57.691 KJ/mol

So is this right then? and Rabn i don't understant where you got (.9*Cp water) from...

[karina]

I think the q value i need to use is the (2884.5 +311.5) for the final calcualtion that is to calculate the neuralization per mole..