# How to calculate the temperature from heat of neutralisation?

The heat of neutralization of $\ce{HCl~ (aq)}$ by $\ce{NaOH}$ is $\mathrm{-55.9~kJ/mol}$ $\ce{H2O}$ produced. If $\mathrm{50~ml}$ of $\mathrm{1.6~M}$ $\ce{NaOH}$ at 25.15 celsius is added to $\mathrm{25~ml}$ of $\mathrm{1.79~M}$ $\ce{HCl}$ at 26.34 celsius in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occurred?

I'm always overwhelmed by the question if it provided me too much information. I don't know what equation and concept to use first. (But I know all related-equations in this unit.)

I need a detailed explanation to understand! Im not sure which one is the initial temperature? 25.15 celsius of $\ce{NaOH}$ or 26.34 celsius $\ce{HCl}$?

• You may get voted down for no obvious attempt to solve the problem. But anyways, here is what I think is the correct concept of the solution to this problems: use the equation q=mC$delta\t$. You know the C is the that of water because it is all about solutions. And for you can find the number of moles of each compounds and multiply them by their molecular weight. And finally you have the amount of energy. All you need to do now is to plug them into the formula solve for change in temp. Since you know the initial temp, you can use the temp difference to find the final temp. – most venerable sir Nov 15 '14 at 3:11
• How can the reaction produce hydrogen? Or did you write down the wrong units in the heat of neutralization? – LDC3 Nov 15 '14 at 3:13
• What is AP level? Is it like college or high school? Or somewhere between? – most venerable sir Nov 15 '14 at 3:17
• Are u a teacher? – most venerable sir Nov 17 '14 at 2:49
• @Doeser AP stands for Advanced Placement. An AP course is a course you take in high school, that you may get college credit for if you do well on the exam. It is not required to get into a college or university. So AP chemistry is equivalent to first year college general chemistry. – DavePhD Nov 17 '14 at 20:07

Second, given the amount of limiting reagent, how much heat will be released (or absorbed). $0.025L \times 1.79M \times 55.9 kJ/mol$
If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. So here: $(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$
So finally calculate the amount the temperature of 75 ml of "water" will increase when $0.025L \times 1.79M \times 55.9 kJ/mol$ of heat is added, and add this to 25.55 degrees C.
• If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. As long as there is no phase changes. – LDC3 Nov 18 '14 at 2:20
• So in $Q = m c_p \Delta T$, I can just let $Q$ equal the heat of neutralization? – TMOTTM Aug 27 '15 at 20:00