To prepare a buffer solution of pH = 6 by mix | vedclass

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(B) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $pH = 6$ and $K_a = 10^{-5}$.F

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$10 : 1$

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(B) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $pH = 6$ and $K_a = 10^{-5}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(10^{-5}) = 5$.Substitute the 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 is $10 : 1$.

To prepare a buffer solution of $pH = 6$ by mixing sodium acetate and acetic acid,what should be the ratio of the concentration of salt to acid? $(K_a = 10^{-5})$

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