Question

About 18.5 mL of 0.10 M HCl was reacted with 13.0 mL of 0.14 M NaOH. Each solution was originally at 25.0 °C, and then is thoroughly mixed. The final temperature of the neutralized mixture is 29.7 °C. Find how much heat (J) was lost/gained by the reaction (system).

Answer 1

Ans. Note that there is an increase in the temperature of the whole reaction mixture (not just the acid or the base). So, we need to account the heat change for the total mass of the reaction mixture.

# Part 1: Heat gained or lost? The temperature of reaction mixture (system) increases from 25.0⁰C to 29.7⁰C- the increase in temperature indicates gain of heat by the system.

# Part 2: Heat gained, Q by a system is given by-

            Q = m s dT

                        Where, m= mass of the reaction mixture

                                    s = specific heat of reaction mixture. Since s is not mentioned in the question, we assume it to be equal to that of water.

                                    dT = Increase in temperature = (Final – Initial) temperature

                                                = 29.7⁰C – 25.0⁰C = 4.7⁰C

# Mass of reaction mixture: Only the volume of acid and bases are mentioned, but not their respective densities. It’s assumed that their density is equal to 1.00 g/mL (that of water) at 25.0⁰C. So, their masses in grams are numerically equal to their respective volumes in mL.

            Now, m = 18.5 g + 13.0 g = 31.5 g

# Now, putting the values in above expression-

            Q = 31.5 g x 4.184 g J ⁰C^-1 x 4.7⁰C = 619.44 J