When a 18.0 mL sample of a 0.308 M aqueous hydrofluoric acid solution is titrated with a 0.361 M aqueous sodium hydroxide solution, what is the pH after 23.0 mL of sodium hydroxide have been added?

pH = 12.8

Explanation:

HF + NaOH → F⁻ + Na⁺ + H₂O

1 mole of HF reacts with 1 mole of NaOH

Initial moles of HF and NaOH are:

HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF

NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH

That means moles of NaOH remains after reaction are:

8.303x10⁻³mol - 5.544x10⁻³mol = 2.759x10⁻³moles NaOH

Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L

Molar concentration of NaOH is

2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]

pOH = - log [OH⁻] = 1.17

As pH = 14 - pOH

pH = 12.8

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !