Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
