Question

PLEASE HELP WITH THIS CHEMISTRY QUESTION:

For 550.0 mL of a buffer solution that is 0.170 M in CH3CH2NH2 and 0.150 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Answer 1

Kb for ethylalamine = 5.6 x 10^-4.

pKb = -log( 5.6 x 10^-4.) = 3.25

pOH = 3.25 + log (0.295/0.325)

pOH = 3.207

14 -3.207 = 10.79 for Initial

lets see for final

0.325 * 0.28 = 0.091mol CH3CH2NH2

0.295 * 0.28 = 0.0826 mol CH3CH2NH3Cl

0.0826 - 0.005 = 0.0776mol CH3CH2NH3Cl

0.091 + 0.005 = 0.096 mol CH3CH2NH2

pOH = 3.25 + log(0.0776 / 0.096)

pOH = 3.157

pH =10.84 for final