Hello,

I would like to do 4 % NaOH solution that is 40 degrees. I have got demineralised water that is 30 degrees and if I add 7,2 kg of solid NaOH to 180 L of 30 degree water,

should the exothermic reaction give approximately 40 °C 4% NaOH solution? (So the temperature should rise from 30 to 40 degree celcius?)

Could you give me some basic calculations?

What i have tried myself:

moles of solute = 7200g/(40g/mol) = 180 mol NaOH

molar heat of solution: ΔHsolution/(moles of solute) = -44.2kJ/mol

heat of solution: ΔHsolution = -44.2kJ/mol • 180 mol = -7956kJ

masssolution = 7,2 kg + 180 = 187,2 kg

ΔTsolution = -ΔHsolution/(masssolution • specific heatsolution) ≈

≈ -ΔHsolution/(masssolution • specific heatwater) =

= 7956kJ/(187,2 kg • 4.184kJ/kg°C) = 10,15 °C

So the final temperature would be: 30 °C+ 10,15°C= 40,15°C.

If i am wrong somewhere please guide me with this calculation.

Also the same problem is with Hydrochloric acid solution. I am adding 32,5 % (m.v) HCL to 180 L of 30 degree water to get 3% HCL solution. So I add 16,6L of 32,5 % HCL to 30 °C 180L demineralised water.

How much will the diluted hydrocloric acid solution temperature increases after adding 16,6 L the concentrated HCL to 180 L of demineralised water that temperature is 30 degrees and concentrated HCL (32,5 %) temperature is 23 degrees?

Could you give me some guidance with this calculation?

**Edited by lak1r, 10 April 2023 - 05:59 AM.**