Question

50.0 mL of a solution of HCl is combined with 100 mL of 1.05 NaOH in a calorimeter. The reaction mixture is initially 22.4 degrees C and the final temp is 30.2 degrees C. What is the molarity of the HCl solution?

Assume there is excess of base, (all HCl reacted) and specific heat of reaction mixture is .96 cal/g C and density of reaction mixture is 1.02 g/ml. The neautralization of HCl and NaOH is 13.6 kcal/mole

Answer 1

we have

heat of neutralization of HCl and NaOH = 13.6 kcal/mole.

Moles of NaOH = molarity x volume = 1.05 x 0.100 L = 0.105 Mol/L

mass of solution = ( 100 + 50 ) ml x 1.02 g/ml = 153.0 g

and

heat = q = m s ΔT  = 153.0 g x 0.96 cal/g C x (30.2 - 22.4 ) C

= 1145.664 cal

moles of the reaction = heat / heat of neutralization

= 1145.664 cal / 13.6 kcal/mole.

= 0.08424 mol

therefore

molarity of HCl = 0.08424 mol / 50 ml

= 0.08424 mol / 0.050 L

= 1.6848 M