At 25 degrees Celsius 25.00 ml of 0.500 M NaOH is neutralized by 1.00 HCL. What is the net ionic equation for the neutralization reaction and the ph at the equivalence point?

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I expect you meant to say "1.00 M HCl," but for purposes of this answer it doesn't actually matter.

The first step toward the net ionic equation is to write the full chemical equation. An acid and a base react in a neutralization reaction to form `H_2O` and a salt. The salt is made up of the cation from the base and the anion from the acid; in this case it is NaCl.

`HCl + NaOH -> H_2O + NaCl`

HCl is a strong acid and will be fully ionized at `25^o C` . ` `

NaOH dissociates into ions in solution, `H_2O` is a liquid, and the salt remains dissolved: ` `

`H^+(aq) + Cl^(-) (aq) + Na^+(aq) + OH^(-) (aq) -> H_2O(l) + Na^+(aq) + Cl^(-) (aq)`

To get the net ionic equation, we remove spectator ions from both sides, leaving

`H^+(aq) + OH^(-)(aq) -> H_2O(l)`

At the equivalence point there is no remaining acid or base, only the salt. Since HCl and NaOH are a strong acid and a strong base respectively, the resulting salt, NaCl, is a neutral salt. Thus the pH at the equivalence point is 7.0 at `25^o C` .

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