Question

Calculate the oh- in 0.01 m aqueous solution of naocn

Answer 1

As you may know

Sodium cyanate (NaOCN) is salt of strong base (NaOH) and weak acid (HOCN).

In aqueous solution, NaOCN will ionise and cyanate (OCN-) ions will hydrolyse acting as a proton acceptor while water will act as proton donor.

NaOCN >>> Na+ + OCN-

OCN- +H2O >>> HOCN + OH-

m(1-a) ma ma

a=degree of dissociation

m=concentration

At equilibrium

OH- ions =ma

[OH-] = √(Kbm)

[OH-] = √(10^-10×0.01)

[OH-] = √(10^-12)

[OH-] = 10^-6

pOH = -log[OH-]

pOH = -log(10^-6)

pOH = 6

As you asked [OH-] = 10^-6 m

Answer 2

hope it helps you............