Get expert advice and community support for all your questions on IDNLearn.com. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
Sagot :
To determine the pH of a 0.050 M benzoic acid (HA) solution with a given acid dissociation constant [tex]\(K_a = 6.3 \times 10^{-5}\)[/tex], follow these steps:
1. Expression of Equilibrium:
Benzoic acid dissociates in water according to the following reaction:
[tex]\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \][/tex]
Here, initially, the concentration of HA is 0.050 M, while the initial concentrations of [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{A}^-\)[/tex] are 0 M.
2. Setting up the Equilibrium Concentrations:
At equilibrium, let [tex]\(x\)[/tex] be the concentration of [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{A}^-\)[/tex] ions due to the dissociation of HA. Then the equilibrium concentrations will be:
[tex]\[ [\text{HA}] = 0.050 - x \][/tex]
[tex]\[ [\text{H}^+] = x \][/tex]
[tex]\[ [\text{A}^-] = x \][/tex]
3. Equilibrium Expression using [tex]\(K_a\)[/tex]:
The equilibrium expression for the dissociation can be written as:
[tex]\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \][/tex]
Substituting the equilibrium concentrations into this expression gives:
[tex]\[ 6.3 \times 10^{-5} = \frac{x \cdot x}{0.050 - x} \][/tex]
4. Simplifying the Expression:
Since [tex]\(K_a\)[/tex] is small, we can assume [tex]\(x \ll 0.050\)[/tex], thus [tex]\(0.050 - x \approx 0.050\)[/tex]. This simplifies our equation to:
[tex]\[ 6.3 \times 10^{-5} \approx \frac{x^2}{0.050} \][/tex]
5. Solving for [tex]\(x\)[/tex]:
Now, solve for [tex]\(x\)[/tex] (which represents the [tex]\(\text{H}^+\)[/tex] concentration):
[tex]\[ x^2 \approx 6.3 \times 10^{-5} \times 0.050 \][/tex]
[tex]\[ x^2 \approx 3.15 \times 10^{-6} \][/tex]
[tex]\[ x \approx \sqrt{3.15 \times 10^{-6}} \][/tex]
[tex]\[ x \approx 0.0017748 \][/tex]
Therefore, the concentration of [tex]\(\text{H}^+\)[/tex] ions ([tex]\(x\)[/tex]) at equilibrium is approximately [tex]\(0.0017748\)[/tex] M.
6. Calculating pH:
The pH of the solution is calculated using the formula:
[tex]\[ \text{pH} = -\log [\text{H}^+] \][/tex]
Substituting the value of [tex]\(x\)[/tex]:
[tex]\[ \text{pH} = -\log (0.0017748) \][/tex]
[tex]\[ \text{pH} \approx 2.75 \][/tex]
Therefore, the pH of the 0.050 M benzoic acid solution is approximately [tex]\(2.75\)[/tex].
1. Expression of Equilibrium:
Benzoic acid dissociates in water according to the following reaction:
[tex]\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \][/tex]
Here, initially, the concentration of HA is 0.050 M, while the initial concentrations of [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{A}^-\)[/tex] are 0 M.
2. Setting up the Equilibrium Concentrations:
At equilibrium, let [tex]\(x\)[/tex] be the concentration of [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{A}^-\)[/tex] ions due to the dissociation of HA. Then the equilibrium concentrations will be:
[tex]\[ [\text{HA}] = 0.050 - x \][/tex]
[tex]\[ [\text{H}^+] = x \][/tex]
[tex]\[ [\text{A}^-] = x \][/tex]
3. Equilibrium Expression using [tex]\(K_a\)[/tex]:
The equilibrium expression for the dissociation can be written as:
[tex]\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \][/tex]
Substituting the equilibrium concentrations into this expression gives:
[tex]\[ 6.3 \times 10^{-5} = \frac{x \cdot x}{0.050 - x} \][/tex]
4. Simplifying the Expression:
Since [tex]\(K_a\)[/tex] is small, we can assume [tex]\(x \ll 0.050\)[/tex], thus [tex]\(0.050 - x \approx 0.050\)[/tex]. This simplifies our equation to:
[tex]\[ 6.3 \times 10^{-5} \approx \frac{x^2}{0.050} \][/tex]
5. Solving for [tex]\(x\)[/tex]:
Now, solve for [tex]\(x\)[/tex] (which represents the [tex]\(\text{H}^+\)[/tex] concentration):
[tex]\[ x^2 \approx 6.3 \times 10^{-5} \times 0.050 \][/tex]
[tex]\[ x^2 \approx 3.15 \times 10^{-6} \][/tex]
[tex]\[ x \approx \sqrt{3.15 \times 10^{-6}} \][/tex]
[tex]\[ x \approx 0.0017748 \][/tex]
Therefore, the concentration of [tex]\(\text{H}^+\)[/tex] ions ([tex]\(x\)[/tex]) at equilibrium is approximately [tex]\(0.0017748\)[/tex] M.
6. Calculating pH:
The pH of the solution is calculated using the formula:
[tex]\[ \text{pH} = -\log [\text{H}^+] \][/tex]
Substituting the value of [tex]\(x\)[/tex]:
[tex]\[ \text{pH} = -\log (0.0017748) \][/tex]
[tex]\[ \text{pH} \approx 2.75 \][/tex]
Therefore, the pH of the 0.050 M benzoic acid solution is approximately [tex]\(2.75\)[/tex].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.
