For the diprotic weak acid H2A, Ka1 = 3.0 x 10-6 and K 2 = 9.0 x 10-9 What is the pH of a 0.0450 M solution of H,A?…
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Question “For the diprotic weak acid H2A, Ka1 = 3.0 x 10-6 and K 2 = 9.0 x 10-9 What is the pH of a 0.0450 M solution of H,A?…”
For the diprotic weak acid H2A, Ka1 = 3.0 x 10-6 and K 2 = 9.0 x 10-9 What is the pH of a 0.0450 M solution of H,A? pH = What are the equilibrium concentrations of H, A and A2- in this solution? [H2A) = [A2-) =
Answer
As follows:
For the first dissociation, the equilibrium constant can be expressed as
We can now create the following ICE diagram to calculate equilibrium concentrations. The starting concentration of acid is 0.0450M.
Initial, M
0.0450
0
0
Change is M
-x
+x
+x
Equilibrium and M
0.0450-x
x
x
The Ka1 expression can also be found on the ICE table.
The equilibrium concentrations of the different species following first dissociation are thus
We can now write the second dissociation using the equilibrium concentrations above as starting concentrations:
To calculate equilibrium concentrations for different species following second dissociation, the following equilibrium chart can be created.
Initial, M
0
Change is M
-y
+y
+y
Equilibrium and M
y
We can thus write the expression Ka2 as follows:
The equilibrium concentrations for each species are thus determined after second dissociation.
To answer your question, you can calculate the pH of the solution as follows:
Hence, the pH of the solution is 3.44
approximately.
As you can see,
the equilibrium concentrations are
All answers are combined up to 3 significant numbers.
H2A(aq), + Haq) & Hag
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HA](eq), x H Kai = e, – 3.0×10-6 [H2A] (eq),
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NA-
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-= 3.0×10-6 [HA-leg] [H+](eq] Kai = xX2 [H24] 0.0450 – = 2 = 3.0×10-6 x (0.0450-1) = 3,66×10-4 M
H2A = 0.0445 M – I = 0.0450M – 3.66×10-4 M –0.0446M
HA = (+1=I = 3.66 x 10-4) M
H Asaq (- Haq) + Acryl
NA-
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3.66 x 10-4)
3.66 x 10-4)
3.66 x 10-4″ – y
3.66 x 10-4) + 4
[A’ leg) [H+leg) Ka2 = [HA-(eg) -y 9.0 x 10-9 M yx (3.66 x 10-4 + y) 3.66 x 10-4 – y = 9.0 x 10-9
y<<3.66x10-4
HAJ) = 3.66×10-4 M -y3.66×10-4 M
H+1 = 3.66×10-4 M +y 3.66×10-4 M
42-1 = y = 9.00 x 10-9 M
pH = -log[H+1=-log(3.66x 10-4) = 3.436 – 3.44
H2A = 0.0446M
A-1 = 9.00 x10-9 M
Conclusion
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