Question

Determine the pH of an HNO2 solution of each concentration. In which cases can you not make the simplifying assumption that x is small?

a. 0.250 M
b. 0.0500 M
c. 0.0250 M

Answer 1

a) Given,

[HNO₂] = 0.250 M

We know,

K_a for HNO₂ = 4.5 x 10^-4

The equilibrium reaction for HNO₂ is,

HNO₂(aq) + H₂O(l) NO₂^-(aq) + H₃O⁺(aq)

Drawing ICE chart,

	HNO2(aq)	NO2-(aq)	H3O+(aq)
I(M)	0.250	0	0
C(M)	-x	+x	+x
E(M)	0.250-x	x	x

Now, the K_a expression is,

K_a = [NO₂^-] [H₃O⁺] / [HNO₂]

4.5 x 10^-4 = [x] [x] / [0.250-x]

4.5 x 10^-4 = [x]² / [0.250] -------Here., [0.250-x] 0.250 since, x<<< 0.250

x = 0.0106

[ Here, (0.0106 / 0.250) x 100 = 4.24 % < 5 % , Thus, here we can make the assumption that x is small]

Now, [H₃O⁺] = x = 0.0106 M

We know, the formula to calculate pH,

pH = -log[H₃O⁺]

pH = -log[0.0106]

pH = 1.97

b) Given,

[HNO₂] = 0.0500 M

We know,

K_a for HNO₂ = 4.5 x 10^-4

The equilibrium reaction for HNO₂ is,

HNO₂(aq) + H₂O(l) NO₂^-(aq) + H₃O⁺(aq)

Drawing ICE chart,

	HNO2(aq)	NO2-(aq)	H3O+(aq)
I(M)	0.0500	0	0
C(M)	-x	+x	+x
E(M)	0.0500-x	x	x

Now, the K_a expression is,

K_a = [NO₂^-] [H₃O⁺] / [HNO₂]

4.5 x 10^-4 = [x] [x] / [0.0500-x]

2.25 x 10^-5 - 4.5 x 10^-4x = x²

x² + 4.5 x 10^-4x - 2.25 x 10^-5 = 0

Solving the quadratic equation,

x = 0.004524

[ Here, if we assumed "x" is small and calculated the value of "x", then (0.00474 / 0.05) x 100 = 9.49 % > 5 % , Thus, here we can not make the assumption that x is small]

Now, [H₃O⁺] = x = 0.004524 M

We know, the formula to calculate pH,

pH = -log[H₃O⁺]

pH = -log[0.004524]

pH = 2.34

c) Given,

[HNO₂] = 0.0250 M

We know,

K_a for HNO₂ = 4.5 x 10^-4

The equilibrium reaction for HNO₂ is,

HNO₂(aq) + H₂O(l) NO₂^-(aq) + H₃O⁺(aq)

Drawing ICE chart,

	HNO2(aq)	NO2-(aq)	H3O+(aq)
I(M)	0.0250	0	0
C(M)	-x	+x	+x
E(M)	0.0250-x	x	x

Now, the K_a expression is,

K_a = [NO₂^-] [H₃O⁺] / [HNO₂]

4.5 x 10^-4 = [x] [x] / [0.0250-x]

1.1125 x 10^-5 - 4.5 x 10^-4x = x²

x² + 4.5 x 10^-4x - 1..1125 x 10^-5 = 0

Solving the quadratic equation,

x = 0.00314

[ Here, if we assumed "x" is small and calculated the value of "x", then (0.003354 / 0.0250) x 100 = 13.4 % > 5 % , Thus, here we can not make the assumption that x is small]

Now, [H₃O⁺] = x = 0.00314 M

We know, the formula to calculate pH,

pH = -log[H₃O⁺]

pH = -log[0.00314]

pH = 2.50