Question

1. the kb of (CH3NH2) is 4.4x10^-4. calculate the pH of a 0.35 M aq solution of CH3NH2

help. i keep getring 1.90. where am i doing my calculations wrong?

2. the ka of (HClO) is 3.0 x 10 ^-8. what is the pH oh a 0.0385 m solution of HClO. .

I am coming up with 3.4 but that doesnt seem correct?

Answer 1

CH3NH2 + H2O <-> CH3NH3+ + OH-

KB = [CH3NH3+][OH-]/[CH3NH2]

4.4*10^-4= x*x/(0.35-x)

x = 0.01219

then

[OH-] = 0.01219

pOH = -log(0.01219 =1.9139

this is pOH we need pH so

ph = 14-1.9139= 12.0861

You are doing the correct procedure, just dont forget that x = _OH- and not H+

2)

for

HClO = 3*!0^-8

this is too low, we must include auto protonation of water

2)

KA = [H+][A-]/[HA]

3*10^-8 = x*x/(0.0375-x)

x = sqrt((3*10^-8*0.0375)) = 0.00003354101

[H+] = 0.00003354101 p

pH = -log(0.00003354101 = 4.47442