[hod357]

Today, I finished an Acid-Base titration--titrating H2SO4 (Sulfuric Acid) into a flask with 25 ml of NaOH (Sodium Hydroxide) and 3 drops of Phenolthaline indicator in it. Titrating the Sulfuric acid into the soduim hydroxide, until one drop of Sulfuric acid turned the solution clear.(neutralizing the solution)
--First Problem--
*I am unsure about the chemical equation I have come up*
NaOH + H₂SO₄  --> H₃O + NaSO₄^-2
Although, after i looked at it i felt as though there is no such thing as NaSO₄^-2
I have come up with another one after this, but am still unsure
2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

--Second Problem--
I am still unsure about the chemical equation so i can not be sure about my mole to mole ratio that I used to find the Molarity of H₂SO₄
After I know both Molarities/Concentrations of both the sodium hydroxide and the sulfuric acid, I am very clueless as to how I would be able to determine the [H₃O⁺] and the pH for the Sulfuric acid which I titrated into the Sodium hydroxide.


Data/Facts from the lab so far:
*Average ml of sulfuric acid needed to neutralize the solution = 20.66 ml
*Molarity of the NaOH (Sodium Hydroxide) = .113M


Help would be much appreciated.

[constant thinker]

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

That is correct.

[hod357]

Much thanks. any idea on how to find the [H₃O⁺] or the pH of the Sulfuric acid?

[UnintentionalChaos]

Well, any acid-base reaction involves the following basic principle:

H₃O⁺ + OH^- -> 2H₂O

Since this reaction occurs between a strong base and a strong acid, I imagine that you can assume the reaction goes to completion (all products). A H₃O⁺ ion results from an acidic proton (hydrogen) on an acid being released. Think about how many the sulfuric acid will release compared to the OH^- released from the NaOH. Look up what molarity means. You can figure out how many moles of NaOH you used up, then relate it to the above information to get the molarity of the sulfuric acid solution. pH is just another way of stating the molarity of H₃O⁺.

[hod357]

well.. correct me if i am wrong.

the molarity of the NaOH is .113 moles per Liter, and the
Molarity of the H₂SO₄ is .068 moles per Liter.
So since the amount of hydronium ions put out is double the amount of hydroxide ions
I should double the .068 moles per liter of H₂SO₄ which = .136
I should take .136 and subtract the .113 of the NaOH? which = .023 = 2.3 x 10^-2  ??
I may be very far off with this

[UnintentionalChaos]

Well, think about this. If a liter of 1 molar sodium hydroxide has 1 mol of NaOH in it, how many moles does 25ml have? Apply this principle to get how many moles of NaOH you neutralized in the experiment. If one molecule of sulfuric acid has two hydronium ions to give up, and one molecule of sodium hydroxide has only one hydroxide, the number of moles of sulfuric acid that were used should be easy to calculate. Do the reverse of the first calculation to find the molarity of the sulfuric acid solution. Molarity of H2SO4 should be very easy to convert to the molarity of the hydronium ions. pH is just plugging the molarity of hydronium ions into an equation which should probably be in your notes somewhere or is probably even on wikipedia.

[hod357]

help. 

[Borek]

This is relatively simple stoichiometry. You alreay have reaction equation. Do you know what this equation means? Please read this stoichiometric calculations introduction, especially explanation how to read information given in reaction equation.

[AWK]

Much thanks. any idea on how to find the [H₃O⁺] or the pH of the Sulfuric acid?

pH calculations of H2SO4 depend on its concentration. For 0.0001 M or more diluted (down to 2x10^-7) H₂SO₄ you can use a complete dissociation of acid, hence approximate concentration of H⁺=2 x concentration of acid .
Fo higher concentrations you should solve a quadratic equation based on K₂ of this acid. For concentration > ~0.05 M an ionic strenght should be taken into account.
K₂=[H⁺][SO₄^2-]/[HSO₄^-]