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Yes, the reaction involving the solid is actually an individual step in the equation of the reaction between the solutions. If you were to add the change in enthalpy of the reaction with the solid NaOh to the change in enthalpy of the other step in the reaction (that's adding water and the NaOh pellets) you would find the sum equivalent to the change in enthalpy of the reaction involving the two solutions (this is supported by Hess's law). I suggest that you consider Hess's law for more information.

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17y ago

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Related Questions

What 2 substances are formed when hydrochloric acid solution neutralizes sodium hydroxide solution?

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To determine the volume of potassium hydroxide solution needed to neutralize the hydrochloric acid solution, you can use the formula M1V1 = M2V2. By plugging in the given values, you can calculate the volume of the potassium hydroxide solution required. In this case, the volume of the 0.152 M potassium hydroxide solution needed to neutralize 10.2 ml of the 0.198 M hydrochloric acid solution would be 7.43 ml.


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HCl + NaOH -----> NaCl + H2O I hope it is correct


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