If I mixed $25.0\ \mathrm{cm^3}$ of $0.350\ \mathrm{mol\space dm^{-3}}$ sodium hydroxide solution with $25.0\ \mathrm{cm^3}$ of $0.350\ \mathrm{mol \space dm^{-3}}$ hydrochloric acid. The temperature rose by $\mathrm{2.50^oC}$. Assume that no heat was lost to the surroundings.

What is the molar enthalpy change for the reaction?

How do I know the mass of water?

  • 2
    $\begingroup$ Assume the density of the resultant solution is the same as that of water. $\endgroup$
    – orthocresol
    Sep 28 '15 at 6:44

Regarding your question about the mass of water, lets calculate how much water is produced by the reaction:

$\pu{25.0 cm^3} = \pu{0.025 dm^3}$
$\pu{0.025 dm^3} \cdot \pu{0.350 mol dm^-3} = \pu{0.00875 mol}~\ce{H2O}$
$\pu{0.00875 mol}~\ce{H2O} \cdot \pu{18 g mol^{-1}} = \pu{0.16 g}~\ce{H2O}$

As this only represents a fraction of a percent of the total water mass (50 g), we can pretty safely neglect any water mass contributed from the neutralization reaction. This is justified as the problem instructs us to make assumptions that would introduce similar errors, like assuming no heat is lost to the surroundings (i.e. we're ignoring the calorimeter constant).

From here the problem is straight forward. Because this is a 1:1 molar neutralization reaction, we can just calculate the heat of the reaction, $q$, then divide by the moles of water produced by the neutralization.

So first we calculate the total heat of the reaction from:

$$q=mC_g\Delta T$$ where:
$q=$ heat of the reaction
$m=$ mass of the solution (water) $= \pu{50.0 g}$
$C_g=$ specific heat of the solution (water) $=\pu{4.18 J g^-1 K^-1}$
$\Delta T=$ temperature increase for the reaction $=\pu{2.50 K}$

$$q = \pu{50.0 g} \cdot \pu{4.18 J g^{-1} K^{-1}} \cdot \pu{2.50 K} = \pu{523 J}$$

Then to calculate the molar heat of the reaction we just divide by the number of moles of water produced as calculated earlier:

$$\pu{523 J}/\pu{0.00875 mol} = \pu{59.8 kJ mol^-1}$$


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