Question

For a solution 0.10M of KOCN:

a. Write the hydrolisis reaction.

b. Write the equilibrium expression.

c. Find [H3O+], [OH-], pH, pOH, and [OCN-] for a solution 0-10 of potassium cianide.

Answer 1

a. KOCN (aq) + H2O (l)  $\to$ HOCN (aq) + KOH (aq)

Or, OCN- (aq) + H2O (l)  $\to$ HOCN (aq) + OH- (aq)

b. Kh =  $\frac{[HOCN][OH^{-}]}{[OCN^{-}][H_2O]}$

The concentration of H2O is very large and reagarded as constant.

Now, Kh = $\frac{[HOCN][OH^{-}]}{[OCN^{-}]}$

c.

Concentration	KOCN	HOCN	OH
Initial	C	0	0
Change	-Cx	+Cx	+Cx
Equilibrium	C - Cx	Cx	Cx

Kh = (Cx × Cx) /(C - Cx )

Now, Kh = Cx2/1-x

Or, 1 >> x , then, Kh = Cx2

or, x = √(Kh/C)

Or, Cx = √(Kh× C)

Now, Kh = $\frac{Kw}{Ka}$

From ICE table ; [OH-] = Cx

Or , [OH-]  = ($\sqrt{}\frac{Kw \times C}{Ka}$)

Or, - log [OH-] = $\frac{1}{2}$ ( - logKw + logKa + logC )

Or, pOH = $\frac{1}{2}$ ( pKw - pKa - logC)

Or, pOH = 7 -  $\frac{1}{2}$ (pKa + logC)

pH = 14 - {7 - $\frac{1}{2}$ ( pKa + logC)}

Or, pH = 7 + $\frac{1}{2}$ ( pKa + logC).

Given, C = 0.10 M

pKa of HOCN = 3.46

Then, pH = 7 + $\frac{1}{2}$ ( 3.46 + log 0.1)

= 7 + $\frac{1}{2}$ ( 3.46 -1)

= 8.23.

[H3O+] = 10-8.23 = 5.89 × 10-9 M.

Then,

pOH = 14 - 8.23 = 5.77.

Or, [OH-] = 10-5.77 = 1.7×10-6 M