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(A) For a buffer solution containing a weak acid and its salt,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$

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Decreases by $1$

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(A) For a buffer solution containing a weak acid and its salt,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Let the initial ratio be $R_1 = \frac{[Salt]}{[Acid]}$. Then $pH_1 = pK_a + \log(R_1)$.The problem states that the ratio of acid to salt is increased ten-fold,which means the new ratio $\frac{[Acid]}{[Salt]}$ is $10 	imes \frac{[Acid]_{initial}}{[Salt]_{initial}}$.Therefore,the new ratio of salt to acid is $R_2 = \frac{[Salt]}{[Acid]} = \frac{1}{10} 	imes R_1$.The new $pH$ is $pH_2 = pK_a + \log(R_2) = pK_a + \log(\frac{R_1}{10}) = pK_a + \log(R_1) - \log(10)$.Since $\log(10) = 1$,we get $pH_2 = pH_1 - 1$.Thus,the $pH$ decreases by $1$.

In a mixture of a weak acid and its salt,the ratio of concentration of acid to salt is increased ten-fold. The $pH$ of the solution

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