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.

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

the answer is c. nacl