Animated Solution for Chemistry - Chemical Thermodynamics: When 400 mL of 0.2 M ${\text{H}}_2{\text{SO}}_4$ solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is ............... $\times 10^{-2}$ K. (Round off to the nearest integer). [Use : ${\text{H}}^{+}(aq)+{\text{OH}}^{-}(aq)⟶{\text{H}}_2\text{O}$; ${\Delta }_{y}H=-57.1{\text{ kJ mol}}^{-1}$] Specific heat of ${\text{H}}_2\text{O}=4.18{\text{ J K}}^{-1}{\text{ g}}^{-1}$, density of ${\text{H}}_2\text{O}=1.0{\text{ g cm}}^{-3}$. Assume no change in volume of solution on mixing.

Imagine you are standing in a laboratory, holding two beakers. In your left hand, you have a highly concentrated, corrosive solution of sulfuric acid. In your right hand, a slippery, basic solution of sodium hydroxide. When you pour them together, you aren't just mixing liquids; you are initiating a violent, microscopic war between ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions. They crash into each other, forming stable water molecules, and in their wake, they release a massive burst of thermal energy.

Our mission today is to act as chemical detectives. We need to calculate exactly how much the temperature of this mixture will spike. Let's break down the physics and chemistry behind this beautiful phenomenon.

Mixing Acid and Base

Before we can calculate the heat, we need to know exactly how many soldiers we are sending into battle. We start by calculating the moles of our reactive ions.

For the acid, we have $400\text{ mL}$ of $0.2\text{ M }{\text{H}}_2{\text{SO}}_4$. But here is the critical trap where many students stumble: sulfuric acid is a diprotic acid. This means every single molecule of ${\text{H}}_2{\text{SO}}_4$ acts as a double-barreled shotgun, firing two ${\text{H}}^{+}$ ions into the solution.

Therefore, the moles of ${\text{H}}^{+}$ are calculated as:

On the other side, we have $600\text{ mL}$ of $0.1\text{ M NaOH}$. Sodium hydroxide is a simple, monoacidic base, meaning its n-factor is just 1.

Who Runs Out First?

The concept of a limiting reagent is much like a dance party. Imagine you have 160 boys (${\text{H}}^{+}$) and only 60 girls (${\text{OH}}^{-}$). They can only dance in pairs to form water (${\text{H}}_2\text{O}$).

No matter how many extra boys are waiting around, the music stops as soon as the 60 girls are paired up. In our chemical reaction, the ${\text{OH}}^{-}$ ions are the limiting reagent. They dictate the absolute maximum amount of water that can be formed.

Thus, exactly $0.06\text{ mol}$ of water will be produced.

Now, let's talk about the energy. The problem states that the enthalpy of neutralization is ${\Delta }_{y}H=-57.1{\text{ kJ mol}}^{-1}$. The negative sign simply tells us that the reaction is exothermic—it is giving heat away to the universe.

For every single mole of water formed, $57.1\text{ kJ}$ of heat is released. Since we only formed $0.06\text{ mol}$ of water, we scale this down proportionally:

To make our upcoming temperature calculations easier, let's convert this to Joules:

Calorimetry is the mathematical bridge between the invisible world of energy and the visible world of temperature. The heat released ($q$) is absorbed by the entire mass of the solution, causing its temperature to rise ($\Delta T$). The governing equation is:

First, we need the total mass ($m$) of the solution. We mixed $400\text{ mL}$ and $600\text{ mL}$, giving us a total volume of $1000\text{ mL}$. Since the density is $1.0{\text{ g cm}}^{-3}$, the mass is simply $1000\text{ g}$.

The specific heat capacity ($c$) of the solution is given as $4.18{\text{ J K}}^{-1}{\text{ g}}^{-1}$. Now, we substitute everything into our master equation:

Solving for the temperature change:

The question asks for the answer in the format of $X\times 10^{-2}\text{ K}$. Let's adjust our decimal point:

Rounding off to the nearest integer, we arrive at our final, triumphant answer: 82.