Chemistry HELP ASAP

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.

Show all work and explain any assumptions.

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

First, calculate the molarity of the solution:

(1 mol NaOH/40.0 g NaOH) (5.25 g NaOH/2.10 L solution) = 0.0625 mol/L (or 0.0625 M)

Now, since NaOH is a strong base, it dissociates 100% into Na⁺ and OH^-. So the concentration of OH^- = 0.0625 M.

Since K_w = [H₃O⁺][OH^-] then 1.00 x 10^-14 = [H₃O⁺] (6.25 x 10^-2)

Solving we get, [H₃O⁺] = 1.60 x 10^-13 M