Answer:
4.91
Explanation:
From the given information:
The millimoles of aniline = 100* 0.3065 = 30.65
30.65 millimoles HNO3 must be added to reach the equivalence point.
30.65 = V x 0.5774
V = 30.65/0.5774
V = 53.08 mL HNO3 must be added
Thus, total volume = 53.08 + 100 = 153.08 mL
[salt] = 30.65 / 153.08 = 0.20022 M
At equivalence point
:
pOH = 1/2 [pKw + pKb + logC]
pOH = 1/2 [14 + 4.87 + log 0.20022 ]
pOH =9.086
pH = 14 - 9.086
pH = 4.914
pH ≅ 4.91 to 2 decimal places.
