Calculations?

Question

What is the [H] and [OH] in a neutral solution at 50 degrees Celcius when Kw = 5.47x10^-14

Answer 1

Since this is a neutral solution, [H+] = [OH-]

The equilibrium equation is Kw = [H][A-]/[HA]

well, [HA] is the water concentration, as it is the "acid" in this reversible reaction: H2O -> H+ + OH-

At 50C, the density of water is 988.03 grams/liter; the molecular weight of water is 18.0153, so in 1 L of water we have 988.03/18.0153 = 54.84 moles. Thus, [HA] = 54.84M.

From Kw we know 5.47E-14 = [H+][A-]/[HA]
in this case [OH-] = [A-]

so 5.47E-14 * [HA] = [H+]^2

Thus [H+] =[OH-] = 1.732*10^-6

Note: This is NOT the square root of Kw !!!!!! You have to divide by the molarity of the water!

Answer 2

Just take the square root of the number you have quoted. [H+] and [OH-] are equal.