The dissociation constant of a weak acid is 1 | vedclass

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(B) Given,dissociation constant of weak acid $K_a = 1 	imes 10^{-4}.$The $pH$ of the buffer solution is $5.$Using the Henderson-Hasselbalch equation

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$10 : 1$

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(B) Given,dissociation constant of weak acid $K_a = 1 	imes 10^{-4}.$The $pH$ of the buffer solution is $5.$Using the Henderson-Hasselbalch equation for an acidic buffer:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$First,calculate $pK_a$:$pK_a = -\log(K_a) = -\log(1 	imes 10^{-4}) = 4.$Substitute the values into the equation:$5 = 4 + \log \frac{[Salt]}{[Acid]}$$\log \frac{[Salt]}{[Acid]} = 5 - 4 = 1.$Taking the antilog on both sides:$\frac{[Salt]}{[Acid]} = 10^1 = 10.$Therefore,the ratio $[Salt]/[Acid]$ is $10 : 1.$

The dissociation constant of a weak acid is $1 \times 10^{-4}.$ In order to prepare a buffer solution with a $pH = 5,$ the $[Salt]/[Acid]$ ratio should be

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