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!!
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
*****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
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.
Well, it looks like you're in a bit of a pickle! Don't worry, I'm here to help you out with some chemistry humor.
Now, let's tackle this problem step by step, like a tightrope walker handling a balancing act.
First, let's calculate the initial concentrations of CH3NH2 and CH3NH3+ based on the given volumes and molarities. For CH3NH2, we have (70 mL/100 mL) * 0.5 M = 0.35 M. For CH3NH3+, we have (30 mL/100 mL) * 1.0 M = 0.3 M.
Now, it's time to use the ICE table to help us find out what's happening to the components of the buffer.
I - Initial concentration
C - Change in concentration
E - Equilibrium concentration
We know that HCl will react with CH3NH2 to form CH3NH3+, so we can set up the following equation:
CH3NH2 + HCl ⇌ CH3NH3+ + Cl-
Now, let's fill in the ICE table. Since we're adding 15 mL of 0.1 M HCl, the change in concentration for HCl is +0.015 M. Since they react in a 1:1 ratio, the change in concentration for CH3NH2 is -0.015 M, and the change in concentration for CH3NH3+ is +0.015 M.
Therefore, at equilibrium, the concentration of CH3NH2 would be 0.35 M - 0.015 M = 0.335 M. The concentration of CH3NH3+ would be 0.3 M + 0.015 M = 0.315 M.
Now, we need to calculate the concentration of OH- ions using the Kb value for CH3NH2. From the equation, we can see that for every molecule of CH3NH2 reacting, one molecule of CH3NH3+ is formed along with one molecule of Cl-. This means the concentration of OH- ions will be the same as the concentration of CH3NH3+, which is 0.315 M.
Finally, we can use the formula for pH, which is pH = -log10([H+]). Since H+ and OH- ions make water when combined, we can find the concentration of H+ by subtracting the concentration of OH- from the concentration of HCl.
[H+] = [HCl] - [OH-]
[H+] = 0.015 M - 0.315 M
[H+] = -0.3 M
Oops! It looks like we have a negative concentration of H+. That's not possible, even in the wackiest of chemical reactions.
So, my dear chemistry enthusiast, it seems that adding HCl to this buffer will result in a pH that is out of this world! You might want to double-check your calculations or explore other options.
Remember, chemistry is all about balancing equations... and maybe a few circus acts along the way! Good luck!
To calculate the pH of the solution after adding 15 mL of HCl to the buffer, we will use the Henderson-Hasselbalch equation, which is used for calculating the pH of a buffer solution:
pH = pKa + log ([A-] / [HA])
In this equation, pKa is the negative logarithm of the acid dissociation constant (Ka), and [A-] and [HA] are the concentrations of the conjugate base and acid, respectively. In this case, CH3NH2 is the base and CH3NH3+ is its conjugate acid.
First, let's find the concentration of CH3NH2 and CH3NH3+ in the buffer before adding HCl:
For CH3NH2 (base):
Given: Volume of CH3NH2 = 70 mL
Concentration of CH3NH2 = 0.5 M
Using the formula: Concentration (M) = moles / volume (L)
moles of CH3NH2 = concentration x volume
= 0.5 M x (70 mL / 1000 mL/L) (converting mL to L)
= 0.035 moles
Concentration of CH3NH2 = moles / total volume
= 0.035 moles / (70 mL + 30 mL) (total volume of base + conjugate acid)
= 0.035 moles / 0.1 L
= 0.35 M
For CH3NH3+ (conjugate acid):
Given: Volume of CH3NH3+ = 30 mL
Concentration of CH3NH3+ = 1.0 M
moles of CH3NH3+ = concentration x volume
= 1.0 M x (30 mL / 1000 mL/L) (converting mL to L)
= 0.030 moles
Concentration of CH3NH3+ = moles / total volume
= 0.030 moles / (70 mL + 30 mL) (total volume of base + conjugate acid)
= 0.030 moles / 0.1 L
= 0.30 M
Now, let's calculate the concentration of CH3NH2 and CH3NH3+ after adding 15 mL of HCl to the buffer:
Given: Volume of HCl = 15 mL
Concentration of HCl = 0.1 M
moles of HCl = concentration x volume
= 0.1 M x (15 mL / 1000 mL/L) (converting mL to L)
= 0.0015 moles
To determine the new concentrations of CH3NH2 and CH3NH3+ after the reaction, we will use the concept of stoichiometry. HCl is a strong acid, so it will react completely with CH3NH2 to form CH3NH3+:
HCl + CH3NH2 → CH3NH3+ + Cl-
Since the reaction is 1:1, this means that the concentration of CH3NH2 will be reduced by 0.0015 M, and the concentration of CH3NH3+ will increase by 0.0015 M.
New concentration of CH3NH2 = 0.35 M - 0.0015 M
= 0.3485 M
New concentration of CH3NH3+ = 0.30 M + 0.0015 M
= 0.3015 M
Now we can substitute these values into the Henderson-Hasselbalch equation to calculate the pH. But first, we need to determine the pKa of CH3NH2:
pKa = -log(Ka)
= -log(5.2 x 10^-4)
= 3.28
Substituting the values into the Henderson-Hasselbalch equation:
pH = 3.28 + log (0.3015 M / 0.3485 M)
Using the logarithmic identity: log (a / b) = log(a) - log(b), the equation can be rewritten as:
pH = 3.28 + log (0.3015 M) - log(0.3485 M)
Now, you can use a scientific calculator to calculate the logarithms and sum the values to find the pH.