Question

8.00g of NaOH are dissolved in water to make 3.00L of solution. What is the concentration of hydronium ions, [H3O+], in this solution?

Answer 1

Solution

Given data

mass of NaOH = 8.00 g

volume = 3.00 L

[H3O+] = ?

lets first calculate moles of NaOH

formula to calculate moles is as follows

moles = mass in gram / molar mass

moles of NaOH = 8.00 g / 39.997 g per mol

                             = 0.200 mol NaOH

now lets calculate molarity of NaOH

molarity = moles / volume in liter

molarity of NaOH = 0.200 mol / 3.00 L

                                = 0.06667 M

since NaOH and OH- has 1 : ratio therefore concnetration of OH- is same as concnetration of NaOH

therefore [OH-] = 0.06667 M

now lets calculate concentration of H3O+

Kw = [H3O+] [OH-]

Kw = 1*10^-14  

Lets put the values in the formula and solve for [H3O+]

1*10^-14 = [ H3O+] [ 0.06667]

1*10^-14/ 0.06667 = [H3O+]

1.49*10^-13 M = [H3O+]

Therefore concentration of hydronium ion [H3O+] = 1.49*10^-13 M