IDNLearn.com makes it easy to find accurate answers to your questions. Get step-by-step guidance for all your technical questions from our dedicated community members.
Sagot :
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.
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
