how many grams of water can be produced when 65.5 grams of sodium hydroxide react with excess sulfuric acid.
1 answer:
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Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid () and aqueous sodium hydroxide (NaOH)
First step. Need to know how much moles of the substances are present
= 0.0025 mol NaOH
0.003 mol NaOH * / 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH ->
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH