No ICE chart is needed; you just have to look at what is going to happen when the HCl is added.
HCl + CH3NH2 ------> CH3NH3^+Cl^-
So, the addition of HCl will convert some of the CH3NH2 to CH3NH3^+Cl^- and decrease the amount of CH3NH2.
First, figure out the amount of moles of CH3NH2 and CH3NH3^+Cl^- that you presently have:
Molarity=moles/Volume (L)
Molarity*Volume=moles
0.5 M CH3NH2* 0.070L=moles of CH3NH2
30 mL of 1.0 M CH3NH3^+Cl^-*0.030L=moles of CH3NH3^+Cl^-
The addition of HCl will decrease the amount of CH3NH2 present in solution and increase the amount of CH3NH3^+Cl^- present in solution. Solve for the number of HCl moles present in solution:
0.1 M HCl*0.015L=moles of HCl
Now, let y=moles of CH3NH2 left after addition of HCl and x= moles of CH3NH3^+Cl^- after addition of HCl
moles of CH3NH2-moles of HCl= y
and
moles of CH3NH3^+Cl^-+moles of HCl=x
Use this variation of the Henderson-Hasselbalch equation and solve for pH:
pOH=pkb+log[BH/B]
Let
B=x
BH=y
pkb=-log(5.2 x 10-4)
and
pOH=???
Solve for pOH:
After solving for pOH, solve for pH:
pH+pOH=14
pH=14-pOH
What will the pH be if 15 mL of 0.1 M HCl are added to the buffer obtained by mixing 70 mL of 0.5 M CH3NH2 and 30 mL of 1.0 M CH3NH3+Cl-? (Kb for CH3NH2 is: 5.2 x 10-4 )
I'm having a hard time figuring out what to do first. is the ICE table come in handy for this? Can you give me a walkthrough pls? Thank you so much!!
4 answers
*****Correction of last part:
Use this variation of the Henderson-Hasselbalch equation and solve for pH:
pOH=pkb+log[BH/B]
Let
B=y**
BH=x**
pkb=-log(5.2 x 10-4)
and
pOH=???
Solve for pOH:
After solving for pOH, solve for pH:
pH+pOH=14
pH=14-pOH
Use this variation of the Henderson-Hasselbalch equation and solve for pH:
pOH=pkb+log[BH/B]
Let
B=y**
BH=x**
pkb=-log(5.2 x 10-4)
and
pOH=???
Solve for pOH:
After solving for pOH, solve for pH:
pH+pOH=14
pH=14-pOH
qs, don't we need the concentration of what's left when we use the equation? Or are they the same thing with moles?
I'm not sure that I understand your question. We don't need to calculate the molarity because the volumes are the same, and will cancel out. Therefore, you can use mole amounts for each of the species; I hope this answers your question.
The concentration of each species are the following:
(moles of CH3NH2)-(moles of HCl)= y
(moles of CH3NH3^+Cl^-)+(moles of HCl)=x
These are the new concentrations after addition of HCl. The species are either CH3NH2 or CH3NH3^+Cl^-. And I just put everything in parenthesis just in case that you do not understand my notes.
The concentration of each species are the following:
(moles of CH3NH2)-(moles of HCl)= y
(moles of CH3NH3^+Cl^-)+(moles of HCl)=x
These are the new concentrations after addition of HCl. The species are either CH3NH2 or CH3NH3^+Cl^-. And I just put everything in parenthesis just in case that you do not understand my notes.