Chapter 5, Problem 5.131AP

Interpretation Introduction

Interpretation: The final temperature of the solution for the given data is to be calculated.

Concept introduction: The heat lost or coming out of mixed solution $\left(q_{\text{mix,}\text{lost}}\right)$ is equal to heat gained $\left(q_{\text{mix,}\text{gain}}\right)$ by solution.

$-q_{\text{mix,}\text{lost}}=q_{\text{mix,}\text{gain}}$

To determine: The final temperature of the solution for the given data.

Answer to Problem 5.131AP

Solution

The final temperature of the solution for the given data is $\underset{\_}{38.50°C}$.

Explanation of Solution

Explanation

Given

The mass of sodium hydroxide $\left(m_{\text{NaOH}}\right)$ is $5.00\text{g}$.

The mass of potassium hydroxide $\left(m_{\text{KOH}}\right)$ is $4.20\text{g}$.

The volume of water $\left(V_{{\text{H}}_{\text{2}}\text{O}}\right)$ is $150\text{mL}$.

The specific heat of solution $\left(c_{p,\text{soln}}\right)$ is $75.3\text{J}/\left(\text{mol}\cdot °\text{C}\right)$.

The initial temperature $\left(T_{\text{i}}\right)$ of the solution is $23°\text{C}$.

The enthalpy of sodium hydroxide solution $\left(\Delta H_{\text{soln},\text{NaOH}}\right)$ is $-44.3\text{kJ}/\text{mol}$.

The enthalpy of potassium hydroxide solution $\left(\Delta H_{\text{soln},\text{KOH}}\right)$ is $-56.0\text{kJ}/\text{mol}$.

The number of moles of sodium hydroxide $\left(n_{\text{NaOH}}\right)$  is calculated by the formula,

$n_{\text{NaOH}}=\frac{m_{\text{NaOH}}}{M_{\text{NaOH}}}$ (1)

Where,

• $m_{\text{NaOH}}$ is mass of $\text{NaOH}$.
• $M_{\text{NaOH}}$ is the molar mass of $\text{NaOH}$.

The molar mass of $\text{NaOH}$ is $\begin{array}{c}=1\times \text{Na}+1\times \text{O}+1\times \text{H}\\ =1\times 22.99\text{g}/\text{mol}+1\times 15.99\text{g}/\text{mol+}1\times 1.008\text{g}/\text{mol}\\ =22.99\text{g}/\text{mol}+15.99\text{g}/\text{mol}+1.008\text{g}/\text{mol}\\ =39.988\text{g}/\text{mol}\end{array}$

Substitute the values of $m_{\text{NaOH}}$ and $M_{\text{NaOH}}$ in equation (1)

$\begin{array}{c}{n}_{\text{NaOH}}=\frac{5.00\text{g}}{39.988\text{g}/\text{mol}}\\ =0.125\text{mol}\end{array}$

The heat coming out from $\text{NaOH}$ solution $\left(q_{\text{NaOH}}\right)$ is calculated by the formula,

$q_{\text{NaOH}}=n_{\text{NaOH}}\times \Delta H_{\text{soln},\text{NaOH}}$ (2)

Where,

• $n_{\text{NaOH}}$ is number of moles of $\text{NaOH}$.
• $\Delta H_{\text{soln},\text{NaOH}}$ is enthalpy of sodium hydroxide solution.

Substitute the values of $n_{\text{NaOH}}$ and $\Delta H_{\text{soln},\text{NaOH}}$ in equation (2).

$\begin{array}{c}{q}_{\text{NaOH}}=0.125\text{mol}\times -44.3\text{kJ}/\text{mol}\\ =-5.5375\text{kJ}\end{array}$

The number of moles of potassium hydroxide $\left(n_{\text{KOH}}\right)$  is calculated by the formula,

$n_{\text{KOH}}=\frac{m_{\text{KOH}}}{M_{\text{KOH}}}$ (3)

Where,

• $m_{\text{KOH}}$ is mass of $\text{KOH}$.
• $M_{\text{KOH}}$ is the molar mass of $\text{KOH}$.

The molar mass of $\text{KOH}$ is $\begin{array}{c}=1\times \text{K}+1\times \text{O}+1\times \text{H}\\ =1\times 39.098\text{g}/\text{mol}+1\times 15.99\text{g}/\text{mol+}1\times 1.008\text{g}/\text{mol}\\ =39.098\text{g}/\text{mol}+15.99\text{g}/\text{mol}+1.008\text{g}/\text{mol}\\ =56.096\text{g}/\text{mol}\end{array}$

Substitute the values of $m_{\text{KOH}}$ and $M_{\text{KOH}}$ in equation (3).

$\begin{array}{c}{n}_{\text{KOH}}=\frac{4.20\text{g}}{56.096\text{g}/\text{mol}}\\ =0.0748\text{mol}\end{array}$

The heat coming out from $\text{KOH}$ solution $\left(q_{\text{KOH}}\right)$ is calculated by the formula,

$q_{\text{KOH}}=n_{\text{KOH}}\times \Delta H_{\text{soln},\text{KOH}}$ (4)

Where,

• $n_{\text{KOH}}$ is number of moles of $\text{KOH}$.
• $\Delta H_{\text{soln},\text{KOH}}$ is enthalpy of sodium hydroxide solution.

Substitute the values of $n_{\text{KOH}}$ and $\Delta H_{\text{soln},\text{KOH}}$ in equation (4).

$\begin{array}{c}{q}_{\text{KOH}}=0.0748\text{mol}\times -56.0\text{kJ}/\text{mol}\\ =-4.18\text{kJ}\end{array}$

The total heat coming out or lost from mixture of solution $\left(q_{\text{mix,}\text{lost}}\right)$ is expressed as,

$q_{\text{mix,}\text{lost}}=q_{\text{NaOH}}+q_{\text{KOH}}$ (5)

Where,

• $q_{\text{NaOH}}$ is heat coming out from $\text{NaOH}$ solution.
• $q_{\text{KOH}}$ is heat coming out from $\text{KOH}$ solution.

Substitute the values of $q_{\text{NaOH}}$ and $q_{\text{KOH}}$ in equation (5).

$\begin{array}{c}{q}_{\text{mix,}\text{lost}}=-5.5375\text{kJ}+-4.18\text{kJ}\\ =-9.726\text{kJ}\end{array}$

To convert the above value in joules, use conversion factor.

$\begin{array}{c}1\text{kJ}={10}^3\text{J}\\ -9.726\text{kJ}=-9.726\times {10}^3\text{J}\end{array}$

The new $q_{\text{mix,}\text{lost}}$ value in joules $\left(q_{\text{mix,}\text{lost}}\left(\text{J}\right)\right)$ is $-9.726\times {10}^3\text{J}$.

The heat lost or coming out of mixed solution $\left(q_{\text{mix,}\text{lost}}\right)$ is equal to heat gained $\left(q_{\text{mix,}\text{gain}}\right)$ by solution.

$-q_{\text{mix,}\text{lost}}=q_{\text{mix,}\text{gain}}$ (6)

The heat gained $\left(q_{\text{mix,}\text{gain}}\right)$ by solution is calculated by the formula,

$q_{\text{mix,}\text{gain}}=n_{{\text{H}}_{\text{2}}\text{O}}\times c_{p,\text{soln}}\times \left(T_{\text{f}}-T_{\text{i}}\right)$ (7)

Where,

• $n_{{\text{H}}_{\text{2}}\text{O}}$ is the number of moles of water.
• $c_{p,\text{soln}}$ is the specific heat of solution.
• $T_{\text{f}}$ is the final temperature of solution.
• $T_{\text{i}}$ is the initial temperature of solution.

The number of moles of water $\left(n_{{\text{H}}_2\text{O}}\right)$ is calculated by the formula,

$n_{{\text{H}}_{\text{2}}\text{O}}=\frac{{\rho }_{{\text{H}}_{\text{2}}\text{O}}\times V_{{\text{H}}_{\text{2}}\text{O}}}{M_{{\text{H}}_{\text{2}}\text{O}}}$ (8)

Where,

• ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ is the density of water.
• $V_{{\text{H}}_{\text{2}}\text{O}}$ is the volume of water.
• $M_{{\text{H}}_{\text{2}}\text{O}}$ is the molar mass of water.

The density $\left({\rho }_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water is $1\text{g}/\text{mL}$.

The molar mass $\left(M_{{\text{H}}_{\text{2}}\text{O}}\right)$ of water is $\begin{array}{l}=2\times \text{H}+1\times \text{O}\\ =2\times 1.008\text{g}/\text{mol}+1\times 15.99\text{g}/\text{mol}\\ =18.006\text{g}/\text{mol}\end{array}$

Substitute the values of ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$, $V_{{\text{H}}_{\text{2}}\text{O}}$, and $M_{{\text{H}}_{\text{2}}\text{O}}$ in equation (8).

$\begin{array}{c}{n}_{{\text{H}}_{\text{2}}\text{O}}=\frac{1\text{g}/\text{mL}\times 150.00\text{mL}}{18.006\text{g}/\text{mol}}\\ =8.33\text{mol}\end{array}$

Substitute the values of $n_{{\text{H}}_{\text{2}}\text{O}}$, $c_{p,\text{soln}}$ and $T_{\text{i}}$ in equation (7).

$\begin{array}{c}{q}_{\text{mix,}\text{gain}}=8.33\text{mol}\times 75.3\text{J}/\left(\text{mol}\cdot °\text{C}\right)\times \left(T_{\text{f}}-23°\text{C}\right)\\ =627.24\text{J}/°\text{C}\times \left(T_{\text{f}}-23°\text{C}\right)\end{array}$

Substitute the values of $q_{\text{mix,}\text{gain}}$ and $q_{\text{mix,}\text{lost}}\left(\text{J}\right)$ in equation (6).

$\begin{array}{c}-\left(-9.726\times {10}^3\text{J}\right)=627.24\text{J}/°\text{C}\times \left(T_{\text{f}}-23°\text{C}\right)\\ \frac{9.726\times {10}^3\text{J}}{627.24\text{J}/°\text{C}}=\left(T_{\text{f}}-23°\text{C}\right)\\ 15.50°\text{C}=\left(T_{\text{f}}-23°\text{C}\right)\\ T_{\text{f}}=23°\text{C}+15.50°\text{C}\end{array}$

Simplify the above expression.

$T_{\text{f}}=38.50°\text{C}$

Thus, the final temperature of the solution for the given data is $\underset{\_}{38.50°C}$

Conclusion

The final temperature of the solution for the given data is $\underset{\_}{38.50°C}$