Question

 

HA, a weak acid, with Ka=1.0×10−8, also forms the ion HA−2. The reaction is HA(aq)+A−(aq)⇌HA−2(aq) and its K = 4.0.

1-) Calculate the [H+] in a 1.0 M solution of HA.

2-)Calculate the [A−] in a 1.0 M solution of HA

3-)Calculate the [HA−2] in a 1.0 M solution of HA.

answers;

[HA−2] =

1.8×10−4

M

[A−] =

4.5×10−5

M

[H+] =

2.2×10−4

M ,

Answer 1

The reaction is--- HA(aq) + A^-(aq) ---> HA^-2 (aq)

Now we can get different equations such as---

(a) HA(aq) ---> H⁺(aq)  + A^- (aq) ,

Given, K_a = 1 x 10^-8 i.e. [H⁺] [A^-] / [HA] = 1 x 10^-8

(b) HA (aq) + A^-(aq)  ---> HA^2- ,

Given, K = 4 .0 i.e. [(HA^2-)] / ([HA] [A^-]) = 4.0

(c) [HA] + [A^-] + [(HA^2-)] = 1 M (concentration)

(d) [H⁺] = [A^-] + [(HA^2-)]

Now we have here four equations, four unknowns, two equilibrium equations, a mass balance equation, and a charge balance equation. We substitute equation (d) in equation (b)

(e) K = 4 = [(HA^2-)] / ([HA] [A^-]) = ([H⁺] - [A^-]) / ([HA] [A^-]) = [H⁺] / ([HA] [A^-]) - 1 / [HA]

Substitute equation (a) in equation (e)
(f) K = 4 = [H⁺] [A^-] / ([HA] [A^-]²) - 1 / [HA] = K_a / [A^-]² - 1 / [HA]

We solve for [HA] as a function of [A^-]
(g)----> [HA] = 1 / (K_a / [A^-]² - K)

and also we back-substitute equation (g) in equation (b) and solve for [(HA^2-)] as a function of [A^-]
(h)----> [(HA^2-)] = K ([HA] [A^-]) = K [A^-] / (Ka / [A^-]² - K)

Substitute equation (g) and (h) in equation (c) and solve the polynomial expression for [A^-] we get,
Concentration (Co) = 1 M = [HA] + [A^-] + [(HA^2-)] = 1 / (K_a / [A^-]² - K) + [A^-] + K [A^-] / (K_a / [A^-]² - K)

= (1 + K [A^-]) / (K_a / [A^-]² - K) + [A-]

= ([A-]² + K [A^-]³) / (K_a - K [A^-]²) + [A^-]

=>Co K_a - Co K [A^-]² = [A-]² + K [A^-]³ + K_a [A-] - K [A-]³ , Co = concentration

=> Co K_a - Co K [A^-]² = [A-]² + K_a [A^-]

=> [A^-]² (1 + CoK) + [A^-] K_a - CoK_a = 0

=> [A-] = 4.472 x 10^-5 M = 4.5 x 10^-5 M

Now, Substitute [A^-] into equation (g) and (h) and solve for [HA] and [(HA^2-)]

[HA] = 1 / (K_a / [A^-]² - K) = (1 - 22.36 x 10^-5) M

=> [(HA^2-)] = K ([HA] [A^-]) = 17.88 x 10^-5 M = 1.8 x 10^-4 M

Lastly substitute [HA] and [A^-] into equation (a) and solve for [H⁺]

[H⁺] = K_a [HA] / [A^-] = 22.36 x 10^-5 M = 2.2 x 10^-4 M