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Question

(A) Given that the concentration of the base $B^-$ and its conjugate acid $HB$ are equal,i.e.,$[B^-] = [HB]$.
For a basic buffer,the $pOH$ is given b

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given that the concentration of the base $B^-$ and its conjugate acid $HB$ are equal,i.e.,$[B^-] = [HB]$.
For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \left( \frac{[salt]}{[base]} \right)$.
Since $[B^-] = [HB]$,the ratio $\frac{[salt]}{[base]} = 1$,and $\log(1) = 0$.
Therefore,$pOH = pK_b = -\log(K_b)$.
Given $K_b = 10^{-10}$,we have $pOH = -\log(10^{-10}) = 10$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Thus,$pH = 14 - pOH = 14 - 10 = 4$.

For a buffer solution containing equal concentrations of $B^-$ and $HB$,where the $K_b$ of $B$ is $10^{-10}$,what is the $pH$ of the buffer solution?

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