Chapter 9, Problem 9.83PAE

9.83 A student performing a calorimetry experiment combined 100.0 mL of 0.50 M HCl and 100.0 mL of 0.50 M NaOH in a coffee cup calorimeter. Both solutions were initially at 20.0°C, but when the two were mixed, the temperature rose to 23.2°C.

(a) Suppose the experiment is repeated in the same calorimeter but this time using 200 mL of 0.50 M HCl and 200.0 mL of 0.50 M NaOH. Will the $\Delta \text{T}$ observed he greater than, less than, or equal to that in the first experiment, and why?

(b) Suppose that the experiment is repeated once again in the same calorimeter, this time using 100 mL of 1.00 M HCl and 100.0 mL of 1.00 M NaOH. Will the $\Delta \text{T}$ observed he greater than, less than, or equal to that in the first experiment, and why?

Interpretation Introduction

To determine: A student performing a calorimetric experiment combined $100.0mL\text{ of 0}\text{.50}\text{M}\text{HCl and 100}\text{.0}\text{mL of 0}\text{.50}\text{M}\text{NaOH}$ in a coffee cup calorimeter. Both solution were initially at ${20.0}^{0}c$, but when the two were mixed the temperature rose to ${23.2}^{0}C$.

Suppose the experiment is repeated in the some calorimeter but this time using $200mL\text{ of 0}\text{.50}\text{M}\text{HCl and 200}\text{.0}\text{mL of 0}\text{.50}\text{M}\text{NaOH}$. Will the $\Delta \text{T}$ observed be greater than, less than or equal to that in the first experiment and why?

Explanation of Solution

Moles of $HCl$ in the first experiment

$=$

$\text{molarity}\left(M\right)\times$ Volume of solution in L

$\begin{array}{l}\left(\text{Volume of HCl}=\frac{100mL}{1000mL}\times 1L=\frac{1}{10}L\right)\\ =\left(0.50mole/L\right)\times \left(\frac{1}{10}L\right)\\ =0.05mole\end{array}$

Moles of $NaOH$ in the first experiment $=$ molarity (M) $\times$ volume of solution (L)

$\begin{array}{l}=\left(0.50mol/L\right)\times \left(\frac{1}{10}L\right)\\ =0.05mole\end{array}$

In the first experiment $0.05mole\text{ of }HCl$ reacts with $0.05mole$ of $NaOH$.

In second experiment.

Moles of $HCl=$ molarity (M) $\times$ (volume of solution in L)

$\begin{array}{l}=\left(0.50mole/L\right)\times \left(\frac{200mL}{1000mL}\times 1L\right)\\ =0.10mole\end{array}$

Moles of $HCl=$ molarity (M) $\times$ (volume of solution in L)

$\begin{array}{l}=\left(0.50mole/L\right)\times \left(\frac{200mL}{1000mL}\times 1L\right)\\ =0.10mole\end{array}$

In second experiment just double amount of $HCl$ reacts with double amount of $NaOH$. So, heat produced in second reaction is double of heat produced in first experiment. If all substance becomes double (solution, mass of calorimeter) then temperature change should be same but calorimeter is same as first experiment only solution is double so temperature change must be greater than $1^{st}$ experiment.

Interpretation Introduction

To determine: Suppose that the experiment is repeated once again in the same calorimeter this time using

$100mL\text{ or 1}\text{.00}\text{M HCl and 100}\text{.0}\text{}\text{mL}\text{ of 1}\text{.00}\text{M}\text{NaOH}$, will the $\Delta T$ observed be greater than, less than or equal to that in the first experiment and why?

Explanation of Solution

In $1^{st}$ experiment $0.05\text{ mole of HCl}$, reacts completely with $0.05\text{ mole of HCl}$,

Mole of $HCl$ in this experiment

$\begin{array}{l}=\text{molarity}\left(M\right)\times \text{Volume of solution}\left(\text{in L}\right)\\ =\left(1.00\right)\times \left(\frac{100.0mL}{1000mL}\times 1L\right)\\ =0.1mole\end{array}$

Mole of $NaOH$ in this experiment

$\begin{array}{l}=\text{molarity}\left(M\right)\times \text{Volume of solution}\left(\text{in L}\right)\\ =\left(1.00\right)\times \left(\frac{100.0mL}{1000mL}\times 1L\right)\\ =0.1mole\end{array}$

In this experiment amount of $HCl\text{ and NaOH}$ is twice the first experiment so amount of heat generated in this experiment is also two times of first experiment.

But mass of solution and calorimeter is same in both cases so rise of temperature in this case is also two times than $1^{st}$ experiment hence $\Delta T$. Is greater in this experiment as compare to first experiment.

Conclusion

1. In this experiment $\Delta T$ is just more than the $\Delta T$in the first experiment.
2. In this experiment $\Delta T$ is twice of $\Delta T$ in the first experiment.