The pH of a 0.54M solution of boric acid (HBO3) is measured to be 4.75. Calculate the acid dissociation constant K of boric acid. Round your answer to 2 significant digits. K = ☐ ? x10 X D
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![### Calculating the Acid Dissociation Constant (\(K_a\)) of Boric Acid
The problem provides the following information:
- **Solution concentration**: 0.54 M (Molar)
- **Compound**: Boric acid (\(H_3BO_3\))
- **Measured pH**: 4.75
#### Task:
Calculate the acid dissociation constant (\(K_a\)) of boric acid. Round your answer to two significant digits.
#### Step-by-Step Solution:
1. **Determine the concentration of \(H^+\) ions** using the pH value.
\[
[H^+] = 10^{-\text{pH}} = 10^{-4.75}
\]
2. **Express the concentration in scientific notation**.
\[
[H^+] \approx 1.78 \times 10^{-5} \, \text{M}
\]
3. **Set up the expression for \(K_a\)**:
For a weak acid \(HA\) dissociating into \(H^+\) and \(A^-\):
\[
K_a = \frac{[H^+][A^-]}{[HA]}
\]
Given that the concentration of \(H^+\) is equal to the concentration of \(A^-\) at equilibrium and the initial concentration of \(HA\) is reduced by the concentration of \(H^+\):
\[
\left[H_3BO_3\right] \approx [HA] = 0.54 \, \text{M} - [H^+] \approx 0.54 \, \text{M}
\]
4. **Simplify the \(K_a\) expression**:
\[
K_a = \frac{(1.78 \times 10^{-5})^2}{0.54} = \frac{3.17 \times 10^{-10}}{0.54} \approx 5.87 \times 10^{-10}
\]
### Completion:
After rounding to two significant digits, the acid dissociation constant \(K_a\) of boric acid is:
\[ K_a \approx 5.9 \times 10^{-10} \]
Here you should input the value for \(K_a\) as:
\[ K_a = 5.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc60b291-b343-46d1-985f-51e0feae8cbd%2F074e5d3b-dde3-4010-aea6-fb44d0611fc6%2Ffy07k1_processed.png&w=3840&q=75)
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