Question

Calculate the pH of .346 M aqueous solution hydrocyanic acid HCN, Ka=4.0x10^-10 and the equilibrium concentrations of the weak acid and it’s conjugate base

pH=

HCN equilibrium=

CN-=

Calculate the pH of a .0169 M aqueous solution of hydro fluoric acid HF, Ka=6.6x10^-4 and the equilibrium concentrations of the weak acid and it’s conjugate base

pH=

HF equilibrium=

F-=

Answer 1

1)

[HCN] = 0.346 M

K_a of HCN = 4.0 x 10^-10

The equilibrium reaction for HCN is as follows:

HCN(aq) + H₂O(l)   CN^-(aq) + H₃O⁺(aq)

Draw an ICE chart for the equilibrium as follows:

	HCN(aq)	CN-(aq)	H3O+(aq)
I(M)	0.346	0	0
C(M)	-x	+x	+x
E(M)	0.346-x	x	x

The K_a expression for the equilibrium is as follows:

K_a = [CN^-][H₃O⁺]/[HCN]

Substitute the known values as follows:

4.0 x 10^-10 = [x][x]/[0.346-x]

4.0 x 10^-10 = [x][x]/[0.346] -------------Here [0.346-x]0.346, since x<<0.346

x = 1.18 x 10^-5

From the ICE chart,

[x] = [H₃O⁺] = 1.18 x 10^-5 M

The formula to determine the pH value is as follows:

pH = -log[H₃O⁺]

pH = -log[1.18 x 10^-5]

pH = 4.93

[HCN]_eq = [0.346-x] = 0.346-1.18 x 10^-5 = 0.346 M

[CN^-]_eq =[x] = 1.18 x 10^-5 M

Thus,

pH = 4.93

[HCN]_eq = 0.346 M

[CN^-]_eq = 1.18 x 10^-5 M

2)

[HF] = 0.0169 M

K_a of HF = 6.6 x 10^-4

The equilibrium reaction for HF is as follows:

HF(aq) + H₂O(l)   F^-(aq) + H₃O⁺(aq)

Draw an ICE chart for the equilibrium as follows:

	HF(aq)	F-(aq)	H3O+(aq)
I(M)	0.0169	0	0
C(M)	-x	+x	+x
E(M)	0.0169-x	x	x

The K_a expression for the equilibrium is as follows:

K_a = [F^-][H₃O⁺]/[HF]

Substitute the known values as follows:

6.6 x 10^-4 = [x][x]/[0.0169-x]

6.6 x 10^-4 = [x][x]/[0.0169] -------------Here [0.0169-x]0.0169, since x<<0.0169

x = 0.00334

But, According to the 5% rule, [0.00334/0.0169]x100 = 19.7 % > 5%, Thus, the the assumption [0.346-x]0.346 is not valid.

Thus, to find the value of x, the quadratic equation has to solve as follows:

6.6 x 10^-4 = [x][x]/[0.0169-x]

(1.12x10^-5) -(6.6 x 10^-4)x -x² = 0

Thus,

x² + (6.6 x 10^-4)x-(1.12x10^-5) =0

Solve the quadratic equation and determine the value of x as follows:

x = 0.00308

From the ICE chart,

[x] = [H₃O⁺] = 0.00308 M

The formula to determine the pH value is as follows:

pH = -log[H₃O⁺]

pH = -log[0.00308]

pH = 2.51

[HF]_eq = [0.0169-x] = 0.0169-0.00308 = 0.0138 M

[F^-]_eq =[x] = 0.00308 M

Thus,

pH = 2.51

[HF]_eq = 0.0138 M

[F^-]_eq = 0.00308 M