Textbook Problem 6D.5

[Alvin Lai 3J]

Calculate the pH, pOH, and percentage protonation of solute in each of the following aqueous solutions: (a) 0.057 M NH3(aq); (b) 0.162 M NH2OH(aq); (c) 0.35 M (CH3)3N(aq); (d) 0.0073 M codeine, given that the pKa of its conjugate acid is 8.21.

Can someone run down the steps on how to approach (a)? Thank you!

[KatarinaReid_3H]

For (a) you would want to:

1) NH3 + H2O <-> NH4+ + OH- then do an ice table
I 0.057 M 0 0
C -x +x +x
E .057 x x

ignore x since Kb= 1.8e-5 is small
Then do K = x^2/.057 and solve for x which is equal to the concentration of OH-. then do -log[OH-] = pOH and you can find pH from 14-pOH.
Hope you can see the ice box I tried to format above!

[Raashi Chaudhari 3B]

To solve for the percent protonation I believe that you have to do the ([protonated form]/[initial form]) *100
So, in part a it would be the value of x, (1 x 10^-3) divided by the initial concentration of NH3 (0.0057). Then multiply that quotient by 100 to get your percentage.
Hope this helped!

[705573518]

For (a) you would want to:

1) NH3 + H2O <-> NH4+ + OH- then do an ice table
I 0.057 M 0 0
C -x +x +x
E .057 x x

ignore x since Kb= 1.8e-5 is small
Then do K = x^2/.057 and solve for x which is equal to the concentration of OH-. then do -log[OH-] = pOH and you can find pH from 14-pOH.
Hope you can see the ice box I tried to format above!

why would it not be x^2/ (.057-x) ... why do we not have -x in the denominator??

[Vanessa Wiratmo 3k]

Hello,

When setting up your Ka ratio, you would have -x in the denominator, but since your Ka Value is so small, you can omit the X in the denominator because it will have very little effect on 0.057. 0.057 - a very small x value will still be about 0.057.

[305769107]

For (a) you would want to:

1) NH3 + H2O <-> NH4+ + OH- then do an ice table
I 0.057 M 0 0
C -x +x +x
E .057 x x

ignore x since Kb= 1.8e-5 is small
Then do K = x^2/.057 and solve for x which is equal to the concentration of OH-. then do -log[OH-] = pOH and you can find pH from 14-pOH.
Hope you can see the ice box I tried to format above!

How did you find Kb?

[Karissa_Lee_1A]

For (a) you would want to:

1) NH3 + H2O <-> NH4+ + OH- then do an ice table
I 0.057 M 0 0
C -x +x +x
E .057 x x

ignore x since Kb= 1.8e-5 is small
Then do K = x^2/.057 and solve for x which is equal to the concentration of OH-. then do -log[OH-] = pOH and you can find pH from 14-pOH.
Hope you can see the ice box I tried to format above!

How did you find Kb?

You can find the Kb of NH3 in table 6C.2