Enough of a monoprotic weak acid is dissolved in water to produce a 0.0138 M solution. The pH of the resulting solution 2.46. Calculate the K₂ for the acid. ⒸK₁ = 3.5 ×10-3
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![**Transcription for Educational Website**
Title: Calculating the Acid Dissociation Constant (Kₐ) for a Weak Acid
**Problem Statement:**
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0138 M solution. The pH of the resulting solution is 2.46. Calculate the Kₐ for the acid.
**Given Data:**
- Concentration (M) = 0.0138 M
- pH = 2.46
**Solution:**
To find the acid dissociation constant (Kₐ), we'll use the formula:
\[
K_a = [H^+][A^-]/[HA]
\]
**Step 1: Determine \([H^+]\) from pH:**
\[
[H^+] = 10^{-pH} = 10^{-2.46} = 3.47 \times 10^{-3} M
\]
**Step 2: Calculate \(K_a\):**
Assuming \([A^-]\) is approximately equal to \([H^+]\) in this weak acid solution:
\[
K_a = (3.47 \times 10^{-3})^2/(0.0138 - 3.47 \times 10^{-3})
\]
**Result:**
\[
K_a = 3.5 \times 10^{-3}
\]
**Conclusion:**
The acid dissociation constant \(K_a\) of the monoprotic weak acid is \(3.5 \times 10^{-3}\).
This calculation helps in understanding the strength of weak acids by determining their degree of ionization in water.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00154cd9-82c7-455b-845c-02cf21fdb65c%2F0ab97bc8-5a56-4fe6-92db-b80f4807ff83%2F8q58oz_processed.jpeg&w=3840&q=75)
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