The pH of an aqueous solution of CH_3COONH_4 | vedclass

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(D) For a salt of a weak acid $(wA)$ and a weak base $(wB)$,the $pH$ of the aqueous solution is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_

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(D) For a salt of a weak acid $(wA)$ and a weak base $(wB)$,the $pH$ of the aqueous solution is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.Given: $K_a = 1.0 	imes 10^{-5}$,so $pK_a = -\log(1.0 	imes 10^{-5}) = 5.0$.Given: $pK_b = 4.4$.Substituting these values into the formula: $pH = 7 + \frac{1}{2}(5.0 - 4.4) = 7 + \frac{1}{2}(0.6) = 7 + 0.3 = 7.3$.However,the question states the $pH$ is $6.7$. Since the formula for the $pH$ of a salt of a weak acid and a weak base is independent of the concentration of the salt,the $pH$ remains $7.3$ regardless of the concentration.Therefore,the given $pH$ value of $6.7$ is inconsistent with the provided $pK_a$ and $pK_b$ values,and the concentration cannot be determined from the $pH$.

The $pH$ of an aqueous solution of $CH_3COONH_4$ is found to be $6.7$ at $25\,^{\circ}C$. The concentration of the $CH_3COONH_4$ solution is ($K_a$ for $CH_3COOH = 1.0 \times 10^{-5}$ and $pK_b$ for $NH_4OH = 4.4$)

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