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(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$ Given $K_a = 10^{-5}$,so $pK_a = -\l

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$10: 1$

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(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$ Given $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$. Given $pH = 6$. Substituting these values into the equation: $6 = 5 + \log \frac{[Salt]}{[Acid]}$ $6 - 5 = \log \frac{[Salt]}{[Acid]}$ $1 = \log \frac{[Salt]}{[Acid]}$ Taking the antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^1 = 10$ Therefore,the ratio of salt to acid concentration is $10: 1$.

Addition of sodium hydroxide solution to a weak acid $(HA)$ results in a buffer of $pH \ 6$. If ionisation constant of $HA$ is $10^{-5}$,the ratio of salt to acid concentration in the buffer solution will be

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