Question

An analytical chemist is titrating 111.0 mL of a 0.3700 M solution of aniline (C6H5NH2) with a 0.3500 M solution of HNO3. The pK_b of aniline is 9.37. Calculate the pH of the base solution after the chemist has added 79.1 mL of the HNO_3 solution to it.

Answer 1

Answer:

The answer is "4.31"

Explanation:

aniline millimoles $= 111 \times 0.37 = 41.07$

added $HNO_3$  millimoles $= 79.1 \times 0.35 = 27.685$

$\to 41.07 - 27.685 = 13.385$ millimoles aniline left

$\to 27.685$ millimoles salt formed

total volume$= 111 + 79.1 = 190.1\ mL$

$\to [aniline] = \frac{13.385}{190.1} = 0.07 \ M\\\\\to [salt] =\frac{ 27.685}{ 190.1} = 0.146\ M\\\\\to pOH = pKb + \frac{\log [salt]}{ [base]}\\\\\to pOH = 9.37 + \frac{\log [0.146]}{[0.07]}\\\\\to pOH = 9.69\\\\\to pH = 14 - 9.69\\\\\to pH = 4.31$