Enough of a monoprotic acid is dissolved in water to produce a 1.77 M solution. The pH of the resulting solution is 2.72. Calculate the K₂ for the acid. K₁ =
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![### Question 19 of 45
**Problem:**
Enough of a monoprotic acid is dissolved in water to produce a 1.77 M solution. The pH of the resulting solution is 2.72. Calculate the \(K_a\) for the acid.
**Solution:**
\[ K_a = \boxed{\phantom{answer}} \]
*Explanation:*
Here, we are given:
- The concentration of the monoprotic acid solution: 1.77 M
- The pH of the solution: 2.72
To find the acid dissociation constant (\(K_a\)), follow these steps:
1. Use the pH to determine the concentration of hydrogen ions \([H^+]\):
\[ [H^+] = 10^{-\text{pH}} \]
\[ [H^+] = 10^{-2.72} \]
2. Write the dissociation equation for the monoprotic acid (HA):
\[ HA \rightleftharpoons H^+ + A^- \]
3. Set up the expression for the acid dissociation constant (\(K_a\)):
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Since the acid is monoprotic, \([H^+] = [A^-]\).
4. Calculate \([H^+]\) using the given pH:
\[ [H^+] = 10^{-2.72} \approx 1.91 \times 10^{-3} \text{ M} \]
5. Knowing that \([A^-] = [H^+] = 1.91 \times 10^{-3} \text{ M}\) and the initial concentration of acid is 1.77 M, the concentration of un-dissociated acid \([HA]\) is:
\[ [HA] = 1.77 \text{ M} - 1.91 \times 10^{-3} \text{ M} \]
6. Plug these values into the \(K_a\) expression and solve for \(K_a\):
\[ K_a = \frac{(1.91 \times 10^{-3}) (1.91 \times 10^{-3})}{1.77 - 1.91 \times 10^{-3}} \]
\[ K_a \approx \frac{3.65 \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc325aa0-3b4d-4100-824a-15814ab6662e%2Fae90b01f-bdd9-4329-a17a-1f8cb42f137f%2F8uezb1s_processed.jpeg&w=3840&q=75)
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