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Example · Example 15

Q.A buffer solution is prepared by mixing 0.1 M CH3COOH0.1\ \text{M}\ \text{CH}_3\text{COOH} with 0.1 M CH3COONa0.1\ \text{M}\ \text{CH}_3\text{COONa}. Given Ka(CH3COOH)=1.8×10−5K_a(\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5}, calculate the pH\text{pH} of this buffer.

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pKa=−log⁡(1.8×10−5)=5−log⁡(1.8)≈5−0.2553=4.7447\text{p}K_a = -\log(1.8\times10^{-5}) = 5 - \log(1.8) \approx 5 - 0.2553 = 4.7447. By the Henderson-Hasselbalch equation, pH=pKa+log⁡([salt]/[acid])\text{pH} = \text{p}K_a + \log([\text{salt}]/[\text{acid}]). Here [salt]=[acid]=0.1 M[\text{salt}] = [\text{acid}] = 0.1\ \text{M}, so the ratio is 1 and log⁡(1)=0\log(1) = 0. Therefore pH=4.7447+0≈4.74\text{pH} = 4.7447 + 0 \approx 4.74. [!ANSWER] $\te …

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