300 mL of a 0.694 M HCl aqueous solution is mixed with 300 mL of 0.347 M Ba(OH)2

Question

300 mL of a 0.694 M HCl aqueous solution is mixed with 300 mL of 0.347 M Ba(OH)2 aqueous solution in a coffee-cup calorimeter. Both the solutions have an initial temperature of 28.7 °C. Calculate the final temperature of the resulting solution, given the following information:
H+(aq) + OH- (aq) ? H2O(?) ? ? ? ?Hrxn = -56.2 kJ/mol
Assume that volumes can be added, that the density of the solution is the same as that of water (1.00 g/mL), and the specific heat of the solution is the same as that for pure water, 4.184 J/(gK).

Explanation / Answer

Moles of HCl = volume x molarity

= 0.300 L x 0.694 mol/L

= 0.2082 mol

Moles of Ba(OH)2 = molarity x volume

= 0.347 mol/L x 0.300 L

= 0.1041 mol

The reaction

2HCl + Ba(OH)2 = BaCl2 + 2H2O

2 mol HCl reacts with = 1 Mol Ba(OH)2

0.2082 mol HCl reacts with = 1*0.2082/2 = 0.1041 Mol Ba(OH)2

All moles will be consumed in the reaction

Moles of H2O produced = 0.2082 mol

Total volume of solution = 300 + 300 = 600 mL x 1L/1000 mL

= 0.600 L

Mass of solution = volume x density

= 600 mL x 1g/ml = 600 g

H rxn = mCp(T2-T1)

56.2*1000 J/mol x 0.2082 mol = 600 g x 4.184 J/(gK) x (T2 - 28.7)K

11700.84 = 2510.4 x (T2 - 28.7)

11700.84 = 2510.4 T2 - 72048.48

T2 = 83749.32/2510.4

= 33.36 C

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