Question

What is $ [H^+] $ in mol/L of a solution that is $0.20 \,M$ in $ CH_3COONa $ and $0.10 \,M$ in $ CH_3COOH $ ? $ (K_a\, \, for\, \, CH_3COOH = 1.8 \times 10^{-5}) $

• A. $ 3.5 \times 10^{-4} $
• B. $ 1.1 \times 10^{-5} $
• C. $ 1.8 \times 10^{-5} $
• D. $ 9.0 \times 10^{-6} $

Answer 1

The Correct Option is D

To find the concentration of hydrogen ions $[H^+]$ in a solution of acetic acid (CH_3COOH) and sodium acetate (CH_3COONa), we need to consider the solution as a buffer system. A buffer solution is a mixture of a weak acid and its conjugate base and can maintain pH upon addition of small amounts of acid or base.

In this problem, we're given:

• Concentration of acetic acid, $[CH_3COOH] = 0.10 \, M
• Concentration of sodium acetate, $[CH_3COONa] = 0.20 \, M
• Acid dissociation constant for acetic acid, $K_a = 1.8 \times 10^{-5}$

The equilibrium equation for acetic acid dissociation is:

CH_3COOH \leftrightarrow H^+ + CH_3COO^-

According to the Henderson-Hasselbalch equation, the pH of a buffer solution can be determined by:

pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

where:

• $[A^-]$ is the concentration of the conjugate base, $[CH_3COO^-] = 0.20 \, M$
• $[HA]$ is the concentration of the weak acid, $[CH_3COOH] = 0.10 \, M$
• $pK_a = -\log K_a = -\log(1.8 \times 10^{-5}) \approx 4.74$

Substituting into the equation:

pH = 4.74 + \log\left(\frac{0.20}{0.10}\right) = 4.74 + \log(2)

We know log(2) \approx 0.30, so:

pH = 4.74 + 0.30 = 5.04

Now, to find $[H^+], we use the relation:

pH = -\log [H^+], which implies [H^+] = 10^{-pH}

So the concentration of hydrogen ions is:

[H^+] = 10^{-5.04} \approx 9.0 \times 10^{-6} \, M

Therefore, the correct answer is $9.0 \times 10^{-6}$, which matches the provided correct option.