Buffer solution Questions in English

(NONE) This is a basic buffer solution consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.
The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right)$.
Given $pK_b$ for $NH_4OH$ is $4.74$.
The number of millimoles of base $(NH_4OH)$ = $5 \text{ mL} \times 0.1 \text{ M} = 0.5 \text{ mmol}$.
The number of millimoles of salt $(NH_4Cl)$ = $250 \text{ mL} \times 0.1 \text{ M} = 25 \text{ mmol}$.
The total volume of the solution = $5 \text{ mL} + 250 \text{ mL} = 255 \text{ mL}$.
Since the volume is the same for both,the ratio of concentrations is equal to the ratio of millimoles: $\frac{[Salt]}{[Base]} = \frac{25}{0.5} = 50$.
Now,$pOH = 4.74 + \log(50) = 4.74 + 1.699 \approx 6.44$.
Since $pH + pOH = 14$,we have $pH = 14 - 6.44 = 7.56$.
Thus,the $pH$ is $756 \times 10^{-2}$.

(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 $pH = 14 - pOH = 14 - 10 = 4$.