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Sure! Let's go through the step-by-step solution to find the pH of a 0.0500 M solution of the diprotic weak acid [tex]\(H_2A\)[/tex] with [tex]\(K_{a1} = 3.5 \times 10^{-6}\)[/tex] and [tex]\(K_{a2} = 8.3 \times 10^{-9}\)[/tex].
### Step 1: First Dissociation Equilibrium
The first dissociation of [tex]\(H_2A\)[/tex] can be represented as:
[tex]\[ H_2A \rightleftharpoons H^+ + HA^- \][/tex]
The equilibrium constant expression (Ka1) for this dissociation is:
[tex]\[ K_{a1} = \frac{[H^+][HA^-]}{[H_2A]} \][/tex]
[tex]\[ 3.5 \times 10^{-6} = \frac{[H^+][HA^-]}{[0.0500 - [H^+]]} \][/tex]
Since [tex]\(H_2A\)[/tex] is a weak acid, we assume that [tex]\([H^+] \ll 0.0500\)[/tex]. So, [tex]\([0.0500 - [H^+]]\)[/tex] can be approximated as [tex]\(0.0500\)[/tex].
Rewriting the expression:
[tex]\[ 3.5 \times 10^{-6} = \frac{[H^+]^2}{0.0500} \][/tex]
[tex]\[ [H^+]^2 = 3.5 \times 10^{-6} \times 0.0500 \][/tex]
[tex]\[ [H^+]^2 = 1.75 \times 10^{-7} \][/tex]
[tex]\[ [H^+] = \sqrt{1.75 \times 10^{-7}} \][/tex]
[tex]\[ [H^+] = 1.32 \times 10^{-4} \][/tex]
This [tex]\( [H^+] \)[/tex] value is considered the initial hydrogen ion concentration from the first dissociation.
### Step 2: Second Dissociation Equilibrium
The second dissociation of [tex]\(HA^-\)[/tex] can be represented as:
[tex]\[ HA^- \rightleftharpoons H^+ + A^{2-} \][/tex]
The equilibrium constant expression (Ka2) for this dissociation is:
[tex]\[ K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]} \][/tex]
Using the concentration from the first equilibrium:
[tex]\[ 8.3 \times 10^{-9} = \frac{[H^+][A^{2-}]}{[HA^-]} \][/tex]
From the first dissociation, [tex]\([HA^-] \approx [H^+] = 1.32 \times 10^{-4} \)[/tex] and if we let [tex]\( [H^+] = y \)[/tex] for the second dissociation:
[tex]\[ [HA^-] \approx 1.32 \times 10^{-4} - y \][/tex]
Since [tex]\(K_{a2}\)[/tex] is quite small, we assume [tex]\( [HA^-] \approx 1.32 \times 10^{-4} \)[/tex].
Rewriting the expression:
[tex]\[ 8.3 \times 10^{-9} = \frac{y^2}{1.32 \times 10^{-4}} \][/tex]
[tex]\[ y^2 = 8.3 \times 10^{-9} \times 1.32 \times 10^{-4} \][/tex]
[tex]\[ y^2 = 1.096 \times 10^{-12} \][/tex]
[tex]\[ y = \sqrt{1.096 \times 10^{-12}} \][/tex]
[tex]\[ y = 1.05 \times 10^{-6} \][/tex]
### Step 3: Total [H^+] Concentration
The total [tex]\( [H^+] \)[/tex] concentration is the sum of [tex]\( [H^+] \)[/tex] from both dissociations:
[tex]\[ [H^+]_{total} = 1.32 \times 10^{-4} + 1.05 \times 10^{-6} \][/tex]
[tex]\[ [H^+]_{total} \approx 1.32 \times 10^{-4} \][/tex]
### Step 4: Calculate pH
Finally, calculate the pH:
[tex]\[ \text{pH} = -\log([H^+]_{total}) \][/tex]
[tex]\[ \text{pH} = -\log(1.32 \times 10^{-4}) \][/tex]
[tex]\[ \text{pH} \approx 3.88 \][/tex]
So, the pH of the 0.0500 M solution of [tex]\(H_2A\)[/tex] is approximately 3.88.
### Step 1: First Dissociation Equilibrium
The first dissociation of [tex]\(H_2A\)[/tex] can be represented as:
[tex]\[ H_2A \rightleftharpoons H^+ + HA^- \][/tex]
The equilibrium constant expression (Ka1) for this dissociation is:
[tex]\[ K_{a1} = \frac{[H^+][HA^-]}{[H_2A]} \][/tex]
[tex]\[ 3.5 \times 10^{-6} = \frac{[H^+][HA^-]}{[0.0500 - [H^+]]} \][/tex]
Since [tex]\(H_2A\)[/tex] is a weak acid, we assume that [tex]\([H^+] \ll 0.0500\)[/tex]. So, [tex]\([0.0500 - [H^+]]\)[/tex] can be approximated as [tex]\(0.0500\)[/tex].
Rewriting the expression:
[tex]\[ 3.5 \times 10^{-6} = \frac{[H^+]^2}{0.0500} \][/tex]
[tex]\[ [H^+]^2 = 3.5 \times 10^{-6} \times 0.0500 \][/tex]
[tex]\[ [H^+]^2 = 1.75 \times 10^{-7} \][/tex]
[tex]\[ [H^+] = \sqrt{1.75 \times 10^{-7}} \][/tex]
[tex]\[ [H^+] = 1.32 \times 10^{-4} \][/tex]
This [tex]\( [H^+] \)[/tex] value is considered the initial hydrogen ion concentration from the first dissociation.
### Step 2: Second Dissociation Equilibrium
The second dissociation of [tex]\(HA^-\)[/tex] can be represented as:
[tex]\[ HA^- \rightleftharpoons H^+ + A^{2-} \][/tex]
The equilibrium constant expression (Ka2) for this dissociation is:
[tex]\[ K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]} \][/tex]
Using the concentration from the first equilibrium:
[tex]\[ 8.3 \times 10^{-9} = \frac{[H^+][A^{2-}]}{[HA^-]} \][/tex]
From the first dissociation, [tex]\([HA^-] \approx [H^+] = 1.32 \times 10^{-4} \)[/tex] and if we let [tex]\( [H^+] = y \)[/tex] for the second dissociation:
[tex]\[ [HA^-] \approx 1.32 \times 10^{-4} - y \][/tex]
Since [tex]\(K_{a2}\)[/tex] is quite small, we assume [tex]\( [HA^-] \approx 1.32 \times 10^{-4} \)[/tex].
Rewriting the expression:
[tex]\[ 8.3 \times 10^{-9} = \frac{y^2}{1.32 \times 10^{-4}} \][/tex]
[tex]\[ y^2 = 8.3 \times 10^{-9} \times 1.32 \times 10^{-4} \][/tex]
[tex]\[ y^2 = 1.096 \times 10^{-12} \][/tex]
[tex]\[ y = \sqrt{1.096 \times 10^{-12}} \][/tex]
[tex]\[ y = 1.05 \times 10^{-6} \][/tex]
### Step 3: Total [H^+] Concentration
The total [tex]\( [H^+] \)[/tex] concentration is the sum of [tex]\( [H^+] \)[/tex] from both dissociations:
[tex]\[ [H^+]_{total} = 1.32 \times 10^{-4} + 1.05 \times 10^{-6} \][/tex]
[tex]\[ [H^+]_{total} \approx 1.32 \times 10^{-4} \][/tex]
### Step 4: Calculate pH
Finally, calculate the pH:
[tex]\[ \text{pH} = -\log([H^+]_{total}) \][/tex]
[tex]\[ \text{pH} = -\log(1.32 \times 10^{-4}) \][/tex]
[tex]\[ \text{pH} \approx 3.88 \][/tex]
So, the pH of the 0.0500 M solution of [tex]\(H_2A\)[/tex] is approximately 3.88.
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