(B) Given: $[Salt] = 0.1 \, M$,$[Acid] = 0.1 \, M$ and $K_a = 1.8 \times 10^{-5}$.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[S

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(B) Given: $[Salt] = 0.1 \, M$,$[Acid] = 0.1 \, M$ and $K_a = 1.8 \times 10^{-5}$.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.255 = 4.745$.Substituting the values: $pH = 4.745 + \log \frac{0.1}{0.1} = 4.745 + \log(1) = 4.745 + 0 = 4.745$.Therefore,the $pH$ of the solution is about $4.7$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Buffer solution

Medium

The $pH$ of a simple sodium acetate buffer is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$. If $[Salt] = [Acid] = 0.1 \, M$,the $pH$ of the solution would be about

A
$7$
B
$4.7$
C
$5.3$
D
$1.4$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Buffer solution

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Which of the following mixtures gives a buffer solution with $pH = 9.25$?
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The $pH$ of a solution containing $0.1 \ N$ $NH_4OH$ and $0.1 \ N$ $NH_4Cl$ is $9.25$. What is the $pK_b$ for $NH_4OH$?

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The $pH$ of a simple sodium acetate buffer is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$. If $[Salt] = [Acid] = 0.1 \, M$,the $pH$ of the solution would be about

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The main buffer present in our blood is.....

In some solutions,the concentration of $H_3O^{+}$ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as

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Which of the following mixtures gives a buffer solution with $pH = 9.25$? Given: $pK_{b} (NH_{4}OH) = 4.75$

The $pH$ of a solution containing $0.1 \ N$ $NH_4OH$ and $0.1 \ N$ $NH_4Cl$ is $9.25$. What is the $pK_b$ for $NH_4OH$?

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Which of the following mixtures gives a buffer solution with $pH = 9.25$?
Given: $pK_{b} (NH_{4}OH) = 4.75$