Question

What is the pH of a solution that is composed of 0.20 moles of HCN and 0.035 moles of NaCN dissolved in enough water to make 1.00L? Ka(HCN)= 4.9 x 10^-10.

Answer 1

Given that, a solution is composed of :

Number of moles of HCN = 0.20 moles , number of moles of NaCN = 0.035 moles , the total volume of the solution = 1.00 L and acid dissociation constant of HCN, Ka =

We have to calculate the pH of this solution.

Answer : -

Concentration of a solute (Molarity)

Molarity of HCN in the solution = (0.20 moles) / (1 L)

Molarity (HCN) =  0.20 M

Molarity of NaCN in the solution = (0.035 moles) /(1 L)

Molarity (NaCN) = 0.035 M

​​​​​​NaCN dissociate into ions in the solution as,

The concentration of is,

HCN dissociate into :

Here concentration of HCN, [HCN] = 0.20 M

Concentration of at equilibrium is calculated using ICE table.

	M	M	M
Initial (I)	0.20	0	0.035
Change (C)	-x	+x	+x
Equilibrium (E)	0.20 - x	x	0.035 + x

Dissociation constant,

at equilibrium.

Since 'x' is very small, 0.035+x 0.035 and 0.20-x 0.20

Therefore concentration of H+ at equilibrium , [H+] =M

pH is defined as,

Therefore pH of the solution = 8.55