# Calculate the enthalpy of dissolution in "kJ/mol" of "NaOH"?

## Suppose the specific heat capacity of water is $\text{4.18 J/g"^@ "C}$. We have that $\text{125 g}$ of water experienced an increase in temperature of ${0.121}^{\circ} \text{C}$ when $2.4 \times {10}^{- 4}$ $\text{g}$ of $\text{NaOH}$ was dissolved. (Assume the mass of $\text{NaOH}$ does not change the specific heat capacity of water significantly.)

May 17, 2017

Here's what I got.

#### Explanation:

The idea here is that you can use the heat absorbed by the solution to find the heat given off by the dissolution of the salt.

More specifically, you can assume that

$\Delta {H}_{\text{diss" = -q_"solution}}$

The minus sign is used here because heat lost carries a negative sign.

To find the heat absorbed by the solution, you can use the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat gained by the water
• $m$ is the mass of the water
• $c$ is the specific heat of water
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

As the problem suggests, you can approximate the mass and the specific heat of the solution to be equal to those of the pure water sample.

The temperature increases by ${0.121}^{\circ} \text{C}$, so you know that

$\Delta T = {0.121}^{\circ} \text{C} \to$ positive because the final temperature is higher than the initial temperature

Plug in your values to find

$q = 125 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 0.121 color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{63.22 J}$

So, you know that the solution absorbed $\text{63.22 J}$, which implies that the dissolution of the salt gave off $\text{63.22 J}$. In other words, you have

$\Delta {H}_{\text{diss" = - "63.22 J}}$

Convert the mass of sodium hydroxide to moles by using the compound's molar mass

2.4 * 10^(-4) color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = 6.00 * 10^(-6) $\text{moles NaOH}$

You know that the enthalpy of dissolution when $6.00 \cdot {10}^{- 6}$ moles of sodium hydroxide are dissolved in water, so use this info to find the enthalpy of dissolution when $1$ mole of the salt dissolves

1 color(red)(cancel(color(black)("mole NaOH"))) * (-"63.22 J")/(6.00 * 10^(-6)color(red)(cancel(color(black)("moles NaOH")))) = -1.054 * 10^7 $\text{J}$

Finally, convert this to kilojoules

1.054 * 10^7 color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = 1.054 * 10^4 $\text{kJ}$

Therefore, you can say that the enthalpy of dissolution, or molar enthalpy of dissolution, for sodium hydroxide is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{diss" = - 1.1 * 10^4color(white)(.)"kJ mol}}^{- 1}}}}$

The answer is rounded to two sig figs, the number fo sig figs you have for the mass of sodium hydroxide.

SIDE NOTE The accepted value for the enthalpy of dissolution of sodium hydroxide in water at ${25}^{\circ} \text{C}$ is

$\Delta {H}_{\text{diss" = - "44.51 kJ}}$

http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/enthalpies%20of%20solution%20of%20electrolytes.pdf

As you can see, this is not even close to what we got, so make sure that you double-check the values given to you by the problem.