Question

pH of 0.16M CH3NH2

Answer 1

Ok so I attempted to solve this problem & got it wrong.
I began by setting up an ice table
Kb for CH3NH2 is 4.4*10^(-4) ---> I proceeded to solve for Ka which = 2.27*10^(-11)
Solved for pH and i got 8.28 and it was wrong :|

Answer 2

Can you show more work such as with the ICE table?

Answer 3

Is it CH3NH3+ and not CH3NH2?

Answer 4

1.) Write the balanced Base equation((Leave water out for acid/base equations))
CH3NH2+H2O<-->CH3+OH

2.) Write the equilibrium form, given is Kb=4.4*10^-4

4.4*10^-4= (CH3NH3)(OH)
(CH3NH2)

4.4*10^-4= (0.15M)(OH)
(0.16M)

(OH)= 4.6933*10^-4
3.) Find the pOH
-log(4.6933*10^-4)
pOH=3.33

4.) Find the pH
pH+3.33=14
pH=11

Answer 5

Where did you come up with the numbers for the equilibrium? Particularly, 0.15 M for the concentration of CH3NH3+.

@elaornelas I think you should try using the Kb instead since CH3NH2 is a base. Find the pOH and then convert to pH by: pH + pOH = 14

Answer 6

I did another problem I switched them, it was with acid also .15M but keep for study use,

Work this the same way
1.) CH3NH2+H2O<--> CH3NH3+OH

2.) 4.4*10^-2= (x)^2 Kb= (CH3NH3)(OH)
(0.16-x) (CH3NH2)
OH= .00275
You can neglect the because less than .5%, in chem 2 you can always do this, but in analytic chem you can not, I cancelled out x from both and x on top which left
4.4*10^-2=x
1.16
OH=.00275

3.) -log(.00275)
pOH= 2.61

4.) pH+pOH=14
pH+2.61=14
PH=11.39M

Answer 7

Sorry it has been a while since chem 2

Answer 8

I Divded instead of Multiplied .16 with 4.4*10^-4
so OH=7.04*10^-5
-log(7.04*10)^-5
pOH=4.15
pH+4.15=14
***** pH=9.84M this should be it ******