ASAP PLEASE!!!!

Question

In a solution at 25oC, 5.25 g of sodium hydroxide, NaOH(s), is added to enough water to create a solution of 2.10 L. What is the hydronium ion, H3O+(aq), concentration in this solution

Answer 1

5.25 g NaOH x 1 mol NaOH / 40 g = 0.131 mols NaOH

0.131 mols NaOH / 2.10 L = 0.0625 mols/L = 0.0625 M

Find the [OH-] ...

NaOH ==> Na⁺ + OH^-

0.0625 mol /L = 0.0625 mol / L OH^-

Find [H3O⁺] ...

Kw = 1x10^-14

Kw = [H3O⁺][OH^-]

1x10^-14 = [H3O⁺][0.0625]

[H3O⁺] = 1x10^-14 / 0.0625

[H3O⁺] = 1.6x10^-13 M

(this answer does not take into account the autohydyrolysis of water which would add to the concentration)

It would actually be 1.6x10^-13 M + 1x10^-14 M = 1.7x10^-13 M

Answer 2

The molar mass of NaOH is 40.0g/mol so when you multiply 5.25g*40g/mol = 210mol, so the concentration is 210mol/2.10L = 100mol/L. NaOH is a strong base, so assuming that nearly all of the NaOH dissociates, [OH-] = 100mol/L. Now, the pOH = -log[OH-] = -log100 = -2 and pOH+pH = 14, so pH = 16 and that means [H+] =[H3O+] = 10^-16mol/L.