Calculate the pH after 0.010 mol gaseous HCl is added to 250.0 mL of each of the following buffered solutions:
a. .050 M NH3/.15 M NH4Cl
b. .5 M NH3/ 1.5 M NH4Cl
a. "Well, someone just mixed ammonia and ammonium chloride! Looks like they're having a gas! But let's find out the pH, shall we? For the first solution, we have 0.050 M NH3 and 0.15 M NH4Cl. Now, when HCl, aka the "acid clown," joins the party, it reacts with NH3 to form NH4+. This means that some the NH3 will go "poof!" and turn into NH4+, making our solution more acidic. To determine the pH, we'll calculate the concentration of H+ ions after the reaction. Ready? Let's get this comedy show on the road!"
b. "Time for another ammonia extravaganza! We've got 0.5 M NH3 and 1.5 M NH4Cl. When our clown friend HCl joins the mix, it reacts with NH3 to form NH4+. This means that more NH3 molecules will transform into NH4+, making our solution even more acidic. To find the pH, we'll need to calculate the concentration of H+ ions after this hilarious chemical reaction. Let's dive right in and see what pH level this dynamic duo will bring!"
To calculate the pH after adding gaseous HCl to a buffered solution, we need to determine the reaction that occurs between the HCl and the components of the buffer. In this case, the reaction involves the HCl reacting with the NH3 (ammonia) and NH4+ (ammonium ion) in the buffer.
a. First, let's consider the reaction between HCl and NH3:
HCl + NH3 → NH4+ + Cl-
The HCl reacts with NH3 to form NH4+ and Cl-. This reaction consumes HCl and NH3 while producing NH4+ and Cl- ions. The amount of NH3 consumed will depend on the stoichiometry of the reaction.
The buffer consists of NH3 and NH4Cl, so initially we have 0.050 M NH3 and 0.15 M NH4+ from NH4Cl.
To calculate the pH, we need to determine the new concentrations of NH3 and NH4+ after the reaction occurs. We can do this by considering the stoichiometry of the reaction.
Let's assume x moles of HCl react with x moles of NH3, then further reactions with NH4Cl will result in the production of NH4+ and Cl- ions.
The balanced equation shows a 1:1 stoichiometric ratio between HCl and NH3. Therefore, x moles of HCl will react with x moles of NH3, resulting in the formation of x moles of NH4+ and Cl-.
After the reaction, the final concentrations of NH3 and NH4+ can be calculated by subtracting the moles of HCl reacted from their initial concentrations in the buffer solution.
The initial concentration of NH3 is 0.050 M, and since the stoichiometric ratio is 1:1, the final concentration of NH3 will be 0.050 M - x.
The initial concentration of NH4+ is 0.15 M, and since the stoichiometric ratio is 1:1, the final concentration of NH4+ will be 0.15 M + x.
Knowing the final concentration of NH4+ allows us to calculate the pOH and subsequently the pH of the solution using the following formula:
pOH = -log10 [OH-]
pH = 14 - pOH
b. The same calculations can be applied to the second buffered solution.
The initial concentration of NH3 is 0.5 M, and the initial concentration of NH4+ from NH4Cl is 1.5 M.
Following the steps outlined above, the final concentration of NH3 will be 0.5 M - x, and the final concentration of NH4+ will be 1.5 M + x.
We can calculate the pOH and pH using the same formulas mentioned in part a.
Please note that the exact values of x and the resulting concentrations will depend on the specific amounts of HCl added and the initial buffer concentrations. This calculation assumes complete reaction between the HCl and the components of the buffer.
To calculate the pH after adding a strong acid to a buffered solution, you need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given as follows:
pH = pKa + log([A-]/[HA])
Where:
pH is the desired pH after adding the acid
pKa is the acid dissociation constant of the weak acid in the buffer solution
[A-] is the concentration of the conjugate base in the buffer solution
[HA] is the concentration of the weak acid in the buffer solution
In both cases, we are adding HCl, which is a strong acid, to a buffered solution. This means that the HCl will completely dissociate, and we will have an excess of Cl- ions.
a. Calculate the pH after adding 0.010 mol of HCl to the .050 M NH3/.15 M NH4Cl buffered solution:
1. First, we need to calculate the new concentrations of NH3 and NH4Cl after adding HCl. Since HCl is a strong acid, it reacts completely with NH3 to form NH4+ ions:
0.010 mol HCl reacts with 0.010 mol NH3
Initial concentration of NH3: 0.050 M
Change in concentration of NH3: -0.010 M
Final concentration of NH3: 0.050 M - 0.010 M = 0.040 M
Since NH4+ is not affected by the addition of HCl, the concentration remains the same:
Concentration of NH4Cl: 0.15 M
2. Next, we need to determine the pKa of NH4Cl. The pKa can be found in literature or reference sources. For NH4Cl, the pKa is approximately 9.24.
3. Now we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([A-]/[HA])
pH = 9.24 + log (0.040 M/0.15 M)
Calculating the logarithm: log (0.040/0.15) = -0.795
pH = 9.24 - 0.795 = 8.445
Therefore, the pH after adding 0.010 mol of HCl to the .050 M NH3/.15 M NH4Cl buffered solution is approximately 8.445.
b. Calculate the pH after adding 0.010 mol of HCl to the .5 M NH3/1.5 M NH4Cl buffered solution:
1. We must follow the same process as in part a. The only difference is the initial concentrations of NH3 and NH4Cl:
Initial concentration of NH3: 0.5 M
Initial concentration of NH4Cl: 1.5 M
Change in concentration of NH3: -0.010 M (since 0.010 mol HCl reacts with 0.010 mol NH3)
Final concentration of NH3: 0.5 M - 0.010 M = 0.490 M
Again, the concentration of NH4Cl remains the same at 1.5 M.
2. The pKa for NH4Cl is the same as in part a, 9.24.
3. Apply the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = 9.24 + log (0.490 M/1.5 M)
Calculating the logarithm: log (0.490/1.5) ≈ -0.748
pH = 9.24 - 0.748 = 8.492
Therefore, the pH after adding 0.010 mol of HCl to the .5 M NH3/1.5 M NH4Cl buffered solution is approximately 8.492.
millimols NH3 initial = 250 x 0.05 = 12.5
mm NH4Cl initial = 250 x 0.15 = 37.5
add 0.01 mol = 10 mmols HCl.
.......NH3 + HCl ==> NH4Cl
I......12.5..0.0.....37.5
add..........10..........
C.....-10...-10......+10
E......2.5....0......47.5
Substitute the E line into Henderson-Hasselbalch equation and solve for pH.
b is done the same way.