Consider a 0.10 M solution of a weak polyprotic acid (H 2 A) with the possible v

Question

Consider a 0.10 M solution of a weak polyprotic acid (H 2 A) with the possible values of K a 1 and K a 2 given below. Calculate the contributions to [ H 3 O + ] from each ionization step. Part A K a 1 =1.0× 10 4 ; K a 2 =5.0× 10 5 Express your answers using two significant figures separated by commas. [ H 3 O + ] 1 , [ H 3 O + ] 2 = Part B) I figured this one out. Part C K a 1 =1.0× 10 4 ; K a 2 =1.0× 10 6 Express your answers using two significant figures separated by commas. [ H 3 O + ] 1 , [ H 3 O + ] 2 =

Explanation / Answer

Part A:

H2A (aq)+ H2O (l) <=====> H3O+ *(aq) + HA- (aq) ……(1)

Ka1 = [H3O+]1[HA-]/[H2A] =1.0*10-4

Let x M be the change in concentration of H2A. Therefore, at equilibrium,

[H2A] = (0.1 – x) M

[H3O+]1 = [HA-] = x M

Plug in the values in the expression and write,

1.0*10-4 = (x).(x)/(0.1 – x)

Since Ka1 is small, we need to make an approximation here; x << 0.1 so that we can write (0.1 – x) 0.1

Therefore,

1.0*10-4 = x2/0.1

====> x2 = 1.0*10-4*0.1 = 1.0*10-5

====> x = 3.16*10-3 3.2*10-3

Therefore, [H3O+]1 = 3.2*10-3 M

Now consider the second ionization; we do not have H2A here, rather we have HA- at equilibrium from the first step. [HA-] = [H3O+]1 = 3.2*10-3 M.

Write down the dissociation:

HA- (aq) + H2O (l) <====> H3O+ (aq) + A2- (aq)

Ka2 = [H3O+]2[A2-]/[HA-] = 5.0*10-5

Let y be the change in concentration of HA- now; then [HA-] = (3.2*10-3 – y)

[H3O+]2 = [A2-] = y

Therefore,

5.0*10-5 = (y).(y)/(3.2*10-3 – y)

Small y approximation gives,

y2 = 1.6*10-7

====> y = 4.0*10-4

Therefore, [H3O+]2 = 4.0*10-4 M

Ans: 3.2*10-3 M, 4.0*10-3 M

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.