Calculate the molar ratio of a weak acid HA ( | vedclass

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(B) The Henderson-Hasselbalch equation for an acidic buffer is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $K_a = 10^{-6}$,so $pK_a = -\

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$1$

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(B) The Henderson-Hasselbalch equation for an acidic buffer is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $K_a = 10^{-6}$,so $pK_a = -\log(10^{-6}) = 6$.Given: $pH = 6$.Substituting the values into the equation: $6 = 6 + \log \frac{[Salt]}{[Acid]}$.This simplifies to: $\log \frac{[Salt]}{[Acid]} = 0$.Taking the antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^0 = 1$.Therefore,the molar ratio of the weak acid $[Acid]$ to its salt $[Salt]$ is $\frac{[Acid]}{[Salt]} = \frac{1}{1} = 1$.

Calculate the molar ratio of a weak acid $HA$ $(K_a=10^{-6})$ and its salt with a strong base,so that the $pH$ of the buffer solution is $6$.

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