(A) For a monoacidic base $BOH$,the $pH$ is $9.0$. Therefore,$pOH = 14 - pH = 14 - 9.0 = 5.0$. Since $pOH = -\log[OH^-]$,we have $[OH^-] = 10^{-pOH} =

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(A) For a monoacidic base $BOH$,the $pH$ is $9.0$. Therefore,$pOH = 14 - pH = 14 - 9.0 = 5.0$. Since $pOH = -\log[OH^-]$,we have $[OH^-] = 10^{-pOH} = 10^{-5} \ M$. For a weak base,$[OH^-] = \sqrt{K_b \times C}$,where $C = 0.10 \ M$. Substituting the values: $10^{-5} = \sqrt{K_b \times 0.10}$. Squaring both sides: $10^{-10} = K_b \times 0.10$. $K_b = \frac{10^{-10}}{0.10} = 10^{-9}$. $pK_b = -\log(K_b) = -\log(10^{-9}) = 9.0$. Thus,the values are $1.0 \times 10^{-9}$ and $9.0$.

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

Medium

If the $pH$ of $0.10 \ M$ monoacidic base at $298 \ K$ is $9.0$,the value of $K_{b}$ and $pK_{b}$ at the same temperature are respectively:

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A
$1.0 \times 10^{-9}, 9.0$
B
$1.0 \times 10^{-5}, 5.0$
C
$1.0 \times 10^{-10}, 10.0$
D
$1.0 \times 10^{-4}, 4.0$

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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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If the $pH$ of $0.10 \ M$ monoacidic base at $298 \ K$ is $9.0$,the value of $K_{b}$ and $pK_{b}$ at the same temperature are respectively:

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