A certain weak base has a 𝐾b of 7.80×10−7. What concentration of this base will produce a pH of 10.06?

Answer 1

Let the weak base be represented by B

B + H₂O ==> BH⁺ + OH^-

Kb = [BH⁺][OH^-] / [B]

If the pH = 10.06, then the pOH must = 14 - 10.06 = 3.94

From pOH = 3.94, [OH^-] = 1x10^-3.94 = 1.15x10^-4 M

So, we have

Kb = 7.80x10^-7 = (1.15x10^-4)(1.15x10^-4) / [B]

7.80x10^-7 = 1.32x10^-8 / [B]

[B] = 0.017 M