(B) For a buffer of weak acid $HX$ and its conjugate base $X^-$, the Henderson-Hasselbalch equation for $pOH$ is given by $pOH = pK_b + \log \frac{[X^

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(B) For a buffer of weak acid $HX$ and its conjugate base $X^-$, the Henderson-Hasselbalch equation for $pOH$ is given by $pOH = pK_b + \log \frac{[X^-]}{[HX]}$.Given that the concentrations are equal, $[X^-] = [HX]$, so $\log \frac{[X^-]}{[HX]} = \log(1) = 0$.Therefore, $pOH = pK_b = -\log(K_b) = -\log(10^{-10}) = 10$.Using the relation $pH + pOH = 14$ at $298 \text{ K}$, we get $pH = 14 - pOH = 14 - 10 = 4$.

Class 11 Chemistry 2 Equilibrium: Ionic Equilibrium Buffer solution

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At $298 \text{ K}$, a certain buffer solution contains equal concentrations of $X^-$ and $HX$. The $K_b$ for $X^-$ is $10^{-10}$. What is the $pH$ of this buffer solution?

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A
$10$
B
$4$
C
$2$
D
$6$

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2 Equilibrium: Ionic Equilibrium

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At $298 \text{ K}$, a certain buffer solution contains equal concentrations of $X^-$ and $HX$. The $K_b$ for $X^-$ is $10^{-10}$. What is the $pH$ of this buffer solution?

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What is $[H^{+}]$ of a solution that is $0.01 \ M$ in $HCN$ and $0.02 \ M$ in $NaCN$ ($K_a$ for $HCN = 6.2 \times 10^{-10}$)?

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