# 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?

##### 1 Answer

#### Answer:

The temperature of the solution will increase by

#### Explanation:

**QUICK ANSWER**

The temperature of the solution will increase by **regardless** of how much strong acid and strong base you're mixing because the **enthalpy change of neutralization**, **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

#color(darkgreen)(ul(color(black)(DeltaT_"15 mL" = DeltaT_"5 mL" = 2^@"C")))#

**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

#"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

#"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

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

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

This tells you that when **mole** of a strong acid reacts with **mole** of a strong base to produce **mole** of water, the reaction gives off

So, assuming that the strong acid solution and the strong base solution have concentrations equal to

#((c * 15)/10^3) color(white)(.)"moles HA" " "# and#" " ((c * 15)/10^3)color(white)(.)"moles BOH"#

This means that when you mix

#((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

Similarly, when you mix

#((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

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

#color(blue)(ul(color(black)(q_"absorbed" = m_"sol" * c_"sol" * DeltaT)))#

Here

#q_"absorbed"# is the heat absorbed by the solution#m_"sol"# is themassof the solution#c_"sol"# is thespecific heatof the solution#DeltaT# is thechange in temperature

When you mix **total volume** of the solution will be

#V_"total" = "15 mL" + "15 mL"#

#V_"total" = "30 mL"#

If you take **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

#DeltaT_"15 mL" = ((c * 15 * x)/(30 * rho * c_"sol")) ""^@"C"#

#color(blue)(ul(color(black)(DeltaT_"15 mL" = (1/2 * (c * x)/(rho * c_"sol"))""^@"C")))#

Similarly, you can say that when you mix

#V_"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

#DeltaT_"5 mL" = ((c * 5 * x)/(10 * rho * c_"sol"))""^@"C"#

#color(blue)(ul(color(black)(DeltaT_"5 mL" = (1/2 * (c * x)/(rho * c_"sol"))""^@"C")))#

As you can see, you have

#color(darkgreen)(ul(color(black)(DeltaT_"15 mL" = DeltaT_"5 mL" = 2^@"C")))#