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How to calculate the heat released when sodium hydroxide is dissolved in hydrochloric acid solution? Here is the data I gathered from a lab experiment:

  • $50\ \mathrm{mL}$ of hydrochloric acid solution
  • $2.00\ \mathrm g$ of sodium hydroxide (in solid form)
  • Initial temperature is $24\ \mathrm{^\circ C}$
  • Final temperature is $44\ \mathrm{^\circ C}$

The chemical equation of the reaction is:

$$\ce{NaOH(s) +H+(aq) + Cl- (aq) -> Na+(aq) +Cl- (aq) + H2O(l)}$$

This is the ONLY information I can use and I cannot search up anything online.

I calculated:

$$ \begin{align} q&=mc\Delta T\\ q&=\left(50\ \mathrm g\right)\left(4.18\ \mathrm{J/(g\ ^\circ C)}\right)\left(20\ \mathrm{^\circ C}\right)\\ q&=4180\ \mathrm J \end{align} $$

So calculating the heat released by the reaction, I assumed that the mass is $50\ \mathrm g$ since hydrochloric acid is a solution of hydrogen chloride in WATER and the reaction produces water as well. Therefore, since the density of water is $1\ \mathrm{g/mL}$, $50\ \mathrm{mL}$ must be approximately $50\ \mathrm g$. Also, the specific heat capacity must be similar to water as well which would be about $4.18\ \mathrm{J/(g\ ^\circ C)}$.

So my question is: Is my assumption correct? If not, why?

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    $\begingroup$ I am pretty sure in reality your assumptions are valid. However at a school level, this is the correct and expected way to do this question. But some teachers may insist that you use 52 grams instead of 50 grams since you also add 2 grams of sodium hydroxide. But that depends on the teacher. $\endgroup$
    – Nanoputian
    Dec 14, 2015 at 6:57

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If that's all the information you have (no concentration or mass or density for the HCl solution, no mass for the final solution), and you weren't instructed to obtain any of that information, then what you did (assuming 50g and heat capacity of water) is ok for an introductory lab class.

There will be many sources of error:

The density of HCl solution is greater than water, so the initial solution will be more than 50g.

The heat capacity of HCl solution is less than that of water.

Without knowing the concentration of the initial solution, it is unknown if the final solution is NaCl solution, a solution of HCl and NaCl, or a solution of NaCl and NaOH.

The heat capacity of the final solution will be less than that of water.

The final solution will be somewhat more than 52g.

Just dissolving NaOH in water is exothermic. It can't be assumed that all the NaOH reacts without knowing the concentration of HCl.

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  • $\begingroup$ This answer would have been very useful if it came earlier, but thanks anyways for the useful information. I will take note of it for future lab experiments. $\endgroup$
    – BCooper
    Dec 15, 2015 at 1:26

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