The temperature of a "15-mL" strong acid solution increases by 2^@"C" when "15 mL" of a strong base is added. By how much should the temperature increase if "5 mL" of each are mixed?

Jan 5, 2018

The temperature of the solution will increase by ${2}^{\circ} \text{C}$.

Explanation:

The temperature of the solution will increase by ${2}^{\circ} \text{C}$ regardless of how much strong acid and strong base you're mixing because the enthalpy change of neutralization, $\Delta {H}_{\text{neut}}$, is constant at constant pressure.

Basically, when you increase the quantities of strong acid and strong base that you're mixing, the heat given off and the volume of the resulting solution will increase by the same factor.

"more acid + base" => {("heat given off " uarr color(blue)(" by a factor f")), ("total volume of the solution " uarr color(blue)(" by the same factor f")) :}

The reaction will give off more heat, but there will be more of the resulting solution to heat, so you will end up with

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {T}_{\text{15 mL" = DeltaT_"5 mL" = 2^@"C}}}}}$

$\textcolor{w h i t e}{\frac{a}{a}}$

DETAILED EXPLANATION

The idea here is that the heat given off by the reaction will depend on how many moles of each reactant you mix, i.e. on the volume of each solution, so right from the start, you should expect to have

$\text{heat given off for 5 mL " < " heat given off for 15 mL}$

because you're mixing a smaller number of moles of the strong acid and of the strong base, provided, of course, that the concentrations of the two reactants do not change.

However, you will also have

$\text{volume of solution for 5 mL " < " volume of solution for 15 mL}$

so look for these two factors to cancel each other out and result in the same $\text{2-"""^@"C}$ increase in the temperature of the resulting solution.

A generic thermochemical equation that can describe this neutralization reaction looks like this

${\text{HA"_ ((aq)) + "BOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "BA"_ ((aq))," "DeltaH_"neut" = +xcolor(white)(.)"kJ mol}}^{- 1}$

This tells you that when $1$ mole of a strong acid reacts with $1$ mole of a strong base to produce $1$ mole of water, the reaction gives off $x$ $\text{kJ}$ of heat.

So, assuming that the strong acid solution and the strong base solution have concentrations equal to $c$ $\text{M}$, you can say that you have

$\left(\frac{c \cdot 15}{10} ^ 3\right) \textcolor{w h i t e}{.} \text{moles HA" " }$ and $\text{ " ((c * 15)/10^3)color(white)(.)"moles BOH}$

This means that when you mix $\text{15 mL}$ of each reactant, the reaction will give off

((c * 15)/10^3) color(red)(cancel(color(black)("moles HA"))) * (x color(white)(.)"kJ")/(1color(red)(cancel(color(black)("mole HA")))) = ((c * 15 * x)/10^3)color(white)(.)"kJ"

of heat, the equivalent of $\left(c \cdot 15 \cdot x\right)$ $\text{J}$ of heat.

Similarly, when you mix $\text{5 mL}$ of each reactant, assuming, of course, that their concentrations are still equal to $c$"M", the reaction will give off

((3 * 5)/10^3) color(red)(cancel(color(black)("moles HA"))) * (x color(white)(.)"kJ")/(1color(red)(cancel(color(black)("mole HA")))) = ( (c * 5 * x)/10^3)color(white)(.)"kJ"

of heat, the equivalent of $\left(c \cdot 5 \cdot x\right)$ $\text{J}$ of heat.

So, just as you would expect, the reaction gives of more heat when you're mixing a bigger number of moles of each reactant.

Now, the heat given off by the neutralization reaction will be absorbed by the solution, which is why its temperature increases.

More specifically, you know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{q}_{\text{absorbed" = m_"sol" * c_"sol}} \cdot \Delta T}}}$

Here

• ${q}_{\text{absorbed}}$ is the heat absorbed by the solution
• ${m}_{\text{sol}}$ is the mass of the solution
• ${c}_{\text{sol}}$ is the specific heat of the solution
• $\Delta T$ is the change in temperature

When you mix $\text{15 mL}$ of each reactant, the total volume of the solution will be

${V}_{\text{total" = "15 mL" + "15 mL}}$

${V}_{\text{total" = "30 mL}}$

If you take $\rho$ ${\text{g mL}}^{- 1}$ to be the density of the solution, you can say that the total mass of the solution will be

30 color(red)(cancel(color(black)("mL solution"))) * (rho color(white)(.)"g")/(1color(red)(cancel(color(black)("mL solution")))) = (30 * rho)color(white)(.)"g"

This means that you have

(c * 15 * x) color(red)(cancel(color(black)("J"))) = (30 * rho) color(red)(cancel(color(black)("g"))) * c_"sol" color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("g")))^(-1) ""^@"C"^(-1) * DeltaT_"15 mL"

which gets you

$\Delta {T}_{\text{15 mL" = ((c * 15 * x)/(30 * rho * c_"sol")) ""^@"C}}$

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta {T}_{\text{15 mL" = (1/2 * (c * x)/(rho * c_"sol"))""^@"C}}}}}$

Similarly, you can say that when you mix $\text{5 mL}$ of each reactant, you have

${V}_{\text{total" = "5 mL" + "5 mL" = "10 mL}}$

This time, you will have

(c * 5 * x) color(red)(cancel(color(black)("J"))) = (10 * rho) color(red)(cancel(color(black)("g"))) * c_"sol" color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("g")))^(-1) ""^@"C"^(-1) * DeltaT_"5 mL"#

which gets you

$\Delta {T}_{\text{5 mL" = ((c * 5 * x)/(10 * rho * c_"sol"))""^@"C}}$

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta {T}_{\text{5 mL" = (1/2 * (c * x)/(rho * c_"sol"))""^@"C}}}}}$

As you can see, you have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {T}_{\text{15 mL" = DeltaT_"5 mL" = 2^@"C}}}}}$