Chemistry, 16.04.2021 20:40 homeschool0123
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.
Answers: 3
 millimoles 
millimoles aniline left
millimoles salt formed
![\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\\](/tpl/images/1264/8201/7a097.png)
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An analytical chemist is titrating 111.0 mL of a 0.3700 M solution of aniline (C6H5NH2) with a 0.350...
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