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(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 equat

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(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: $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} ight)$.Given $pK_b$ for $NH_4OH$ is $4.74$.The number of millimoles of base $(NH_4OH)$ = $5 	ext{ mL} 	imes 0.1 	ext{ M} = 0.5 	ext{ mmol}$.The number of millimoles of salt $(NH_4Cl)$ = $250 	ext{ mL} 	imes 0.1 	ext{ M} = 25 	ext{ mmol}$.The total volume of the solution = $5 	ext{ mL} + 250 	ext{ mL} = 255 	ext{ mL}$.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 $756 	imes 10^{-2}$.

The pH of a solution obtained by mixing $5 \text{ mL}$ of $0.1 \text{ M } NH_4OH$ solution with $250 \text{ mL}$ of $0.1 \text{ M } NH_4Cl$ solution is . . . . . . $\times 10^{-2}$.

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