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April 23, 2014

April 23, 2014

Posted by **ogi** on Saturday, March 3, 2012 at 7:08am.

When one mole of KOH is neutralized by sulfuric acid, q=-56kJ. At 22.8 C , 25.0 mL of 0.500 M H2SO4 is neutralized by 50.0 mL of 0.500 M KOH in a coffee-cup calorimeter. What is the final temperature of the solution?

- chemistry -
**DrBob222**, Saturday, March 3, 2012 at 3:57pm2KOH + H2SO4 ==> H2SO4 + 2H2O

q = -56 kJ/mol KOH or 2-56 = -112 kJ/rxn.

50.0 mL x 0.500M KOH = 25 mmols.

25.0 mL x 0.500M H2SO4 = 12.5 mmols.

Total volume = 75 mL and I assume density is considered to be 1.00 g/mL.

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

112 = 75g x 4.184 J/g*K x (Tfinal-22.8)

Solve for Tf

I get an estimate of 23+ C

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