(A) Given: $pH = 5.0$ and concentration $C = 0.01 \ M = 10^{-2} \ M$. $[H^{+}] = 10^{-pH} = 10^{-5} \ M$. For a weak acid,the dissociation constant $K

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$1 \times 10^{-5} \ M, 1 \times 10^{-8}$

(A) Given: $pH = 5.0$ and concentration $C = 0.01 \ M = 10^{-2} \ M$. $[H^{+}] = 10^{-pH} = 10^{-5} \ M$. For a weak acid,the dissociation constant $K_a$ is given by the formula $[H^{+}] = \sqrt{K_a \cdot C}$. Squaring both sides,we get $[H^{+}]^2 = K_a \cdot C$. Therefore,$K_a = \frac{[H^{+}]^2}{C} = \frac{(10^{-5})^2}{10^{-2}} = \frac{10^{-10}}{10^{-2}} = 10^{-8}$. Thus,$[H^{+}] = 1 \times 10^{-5} \ M$ and $K_a = 1 \times 10^{-8}$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of weak Acids and weak Bases

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The $pH$ of $0.01 \ M$ solution of acetic acid is $5.0$. What are the values of $[H^{+}]$ and $K_a$ respectively?

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A
$1 \times 10^{-5} \ M, 1 \times 10^{-8}$
B
$1 \times 10^{-5} \ M, 1 \times 10^{-9}$
C
$1 \times 10^{-4} \ M, 1 \times 10^{-8}$
D
$1 \times 10^{-3} \ M, 1 \times 10^{-8}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of weak Acids and weak Bases

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The $pH$ of $0.01 \ M$ solution of acetic acid is $5.0$. What are the values of $[H^{+}]$ and $K_a$ respectively?

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