[Tymed]

last minute cramming during the holiday break, and I have no clue how to do any of the questions....
just need help with one.
Q: calculate the concentration of hydronium ions in the solution, 18.6 mL of 2.60mol/L ClO4 (aq)added to 24.8 mL of 1.92 mol/L NaOH (aq).

&

Q2: calculate the concentration of hydronium ions in the solution, 17.9mL of 0.175 mol/L HNO3 (aq) added to 35.4 mL of 0.0160 mol/L Ca(OH)2(aq)

thank you !

[UG]

Q: calculate the concentration of hydronium ions in the solution, 18.6 mL of 2.60mol/L ClO4 (aq)added to 24.8 mL of 1.92 mol/L NaOH (aq).

Did you mean HClO₄?
Anyway, I would probably approach this by calculating the number of moles of NaOH and HClO₄. They are a strong base and acid respectively so all of it will dissociate in water to form OH^- and H⁺ ions respectively. When the acid is added to the base, a neutralisation reaction will occur H⁺ + OH^- H₂O
You need to decide which one of OH^- or H⁺ will be limiting and which one is in excess. The limiting reagent will be completely neutralised. You now need to work out the new concentration of the excess ion (since the volume of the solution has changed), from there the concentration of the hydronium ions can be worked out.

[Tymed]

Did you mean HClO₄?

Yes I did mean that, anyways I solved the first one but i can't get the second one more some reason. I found the moles of each individual solution, then found the limiting reagent, then subtracted the excess from the limiting reagent, took that number and divided it by the total volume, and didn't get the right answer.... and the correct answer is, 0.0375 mol/L, but I can't seem to get that answer.

[UG]

Did you take into account that for every unit of Ca(OH)₂, two OH^- ions are produced in water? As such: Ca(OH)₂  Ca²⁺ + 2OH^-