Question

Calculate the pH of the solution that results from the following mixture: 85.0 mL of 0.12 M C2H5NH2 with 270.0 mL of 0.21 M C2H5NH3Cl The Kb of C2H5NH2 is 5.6×10^-4 Please explain in detail.. I don't understand how to get the answer. So far I've gotten 6.66, 3.66, and 7.34 and all of them are wrong.

Answer 1

pH = pKa + ln([A^-]/[HA])

[A^-] = C2H5NH2 (the base)
[HA] = C2H5NH3Cl (the acid)

We can use moles instead of concentrations because the volumes cancel in the concentration ratio ([A^-]/[HA]).

moles C2H5NH2 = (85.0 mL)*(0.12 M C2H5NH2) = 10.2 mmoles C2H5NH2
moles C2H5NH3Cl = (270.0 mL)*(0.21 M C2H5NH3Cl) = 56.7 mmoles C2H5NH3Cl

We are missing the pKa (formally, pKa(H^+)) of C2H5NH3Cl.

pKa (of protonated form) = 10.7

H = pKa + ln([A^-]/[HA])
pH = 10.7 + ln((10.2)/(56.7)) = 10.7 + (-1.7154) = 8.9846

"... pKa (of protonated form) = 10.7 ..."