Question

Calculate the pH of a solution that contains 2.00% (by weight, w/w) NaOH and has a density of 1.022 g/mL.

Please show all work/steps thank you!

Answer 1

mass % = (mass of NaOH/mass of solution)*100

Let mass of solution = 100 g

Thus, mass of NaOH = 2 g

Molar mass of NaOH = 40 g/mole

Thus, moles of NaOH in 2 g of it = mass/molar mass = 2/40 = 0.05

Volume of solution = mass/density = 100/1.022 = 97.847 ml = 0.097847 litres

Thus, Molarity of NaOH = moles of NaOH/volume of solution in litres = 0.05/0.097847 = 0.511 M

Now, NaOH being a strong base dissociates completely into its ions, i.e.

NaOH(aq) ---------> Na⁺(aq) + OH^-(aq)

Thus, [OH^-] = 0.511 M

Thus, pOH = -log[OH^-] = 0.293

Hence pH = 14 - pOH = 13.707