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

The answer is "4.31"

Step-by-step explanation:

aniline millimoles
`= 111 * 0.37 = 41.07`

added
`HNO_3` millimoles
`= 79.1 * 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] = (13.385)/(190.1) = 0.07 \ M\\\\\to [salt] =( 27.685)/( 190.1) = 0.146\ M\\\\\to pOH = pKb + (\log [salt])/( [base])\\\\\to pOH = 9.37 + (\log [0.146])/([0.07])\\\\\to pOH = 9.69\\\\\to pH = 14 - 9.69\\\\\to pH = 4.31\\`