The density of water at room temperature is less than the assumed 1.0 g/mL used in...

The density of water at room temperature is less than the assumed 1.0 g/mL used in the calculations for determining the enthalpy of neutralization. How will this small (but real) assumption affect the reported enthalpy of neutralization for the strong acid-weak base reaction? Explain.

Expert Solution

We make the assumption that strong acids and strong alkalis are fully ionised in solution, and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution.

The equation for any strong acid being neutralised by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction.

The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is:

. . . but what is actually happening is:

If the reaction is the same in each case of a strong acid and a strong alkali, it isn't surprising that the enthalpy change is similar.

queen_honey_blossom answered 4 hours ago

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