Question

A) The heat of neutralization of HCl (aq) by NaOH (aq) is -55.84 kJ/mol H2O produced. 55 mL of 1.72 M NaOH is added to 35 mL of 2.14 M HCl. Both solutions are at 23.10°C.How many moles of water are produced by this reaction? B)The final solution resulting from the reaction in Part A has a density of 1.02 g/mL and a specific heat of 3.98 J/(g·°C). What is the final solution temperature of the solution in Part A?

Answer 1

Ans :

Number of moles of NaOH = molarity x volume (L)

= 1.72 x 0.055 = 0.0946 mol

Number of moles of HCl = 2.14 x 0.035 = 0.0749 mol

Equimolar amounts of both the reagents combine , so the number of moles of water formed will be :

= 0.0749 mol

B)

The heat of neutralization is -55.84 KJ/mol H2O

So here the heat released will be : -55.84 x 0.0749

= -4.182416 KJ

= 4182.416 J

total volume of solution = 55 mL + 35 mL = 90 mL

Density = mass / volume

1.02 = m / 90

m = 91.8 grams

Heat = mass of substance x specific heat x change in temperature

4182.416 = 91.8 x 3.98 x T

T = 11.45^oC

So the final temperature will be 23.10^oC + 11.45^oC

= 34.55^oC