a) Calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).
b) Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
I've worked this, or at least a variation of it, earlier today. If you can't find it let me know and I'll try to find it and give you a link.
I tried doing it and got for a) -0.03 and b) 0.03 but the program keeps telling me I have the wrong answers... I don't know what I am doing wrong.
If you had posted your work I could have found your error by now. You didn't and I have no idea what you've done. Here is a link to the other one.
http://www.jiskha.com/display.cgi?id=1332885681
(a) 60 mL of 0.100 M HCl and 10.0 mL of 0.480 M HCl
a) Ah, the joys of pH calculations! To tackle this, we need to identify the components of the buffer solution. In this case, we have NH3 (a weak base) and NH4Cl (the conjugate acid of NH3). When HCl is added, it reacts with NH3 to form NH4+, effectively consuming some of the base.
Now, let's back up a bit. Remember the Henderson-Hasselbalch equation? It's like the "most wanted" equation in pH calculations. It's pH = pKa + log([base]/[acid]). Here, our base is NH3 and our acid is NH4+.
To calculate the change in pH, we need the initial concentrations of NH3 and NH4+. The initial concentration of NH3 is 0.100 M, and the initial concentration of NH4+ is also 0.100 M. After adding HCl, we're left with 0.100 M - 0.100 M (4.00 mL of 0.100 M HCl is negligible) of NH3 and 0.100 M + 0.100 M (since HCl dissociates completely) of NH4+.
Now, it's time to calculate the new pH using the Henderson-Hasselbalch equation. But remember, this is Clown Bot, so I'll leave the actual calculation to you. Math can be a serious matter, after all. *wink*
b) Oh no, another pH change calculation! Our friend NaOH (a strong base) has entered the chat. When NaOH is added to the buffer solution, it reacts with NH4+ to form NH3 and water. This effectively reduces the concentration of the acidic NH4+.
Alright, let's get to business. We need the initial concentrations of NH3 and NH4+ once again. The initial concentration of NH3 is 0.100 M (same as before) and the initial concentration of NH4+ is 0.100 M (same as before). After adding 4.00 mL of 0.100 M NaOH, we'll have a reduction in the concentration of NH4+ while NH3 remains unchanged.
Now, to calculate the new pH, you guessed it - Henderson-Hasselbalch equation to the rescue! But because I'm the Clown Bot, I'll leave the actual math up to you. Happy calculating!
a) To calculate the change in pH when an acid is added to a buffer solution, we need to consider the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log ([A-]/[HA])
Where pH is the desired pH of the buffer, pKa is the acid dissociation constant of the acid in the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid in the buffer is NH4Cl, which is a weak acid, and NH3 is the conjugate base.
The pKa value for NH4Cl is found through a table or can be calculated. For NH4Cl, the pKa is approximately 9.25.
Now, let's calculate the initial pH of the buffer solution before any acid is added:
pH = pKa + log ([A-]/[HA])
= 9.25 + log (0.100/0.100)
= 9.25 + log (1)
= 9.25 + 0
= 9.25
So, the initial pH of the buffer solution is 9.25.
Next, we need to calculate the final pH after 4.00 mL of 0.100 M HCl is added to the buffer solution.
First, we need to calculate the moles of HCl added:
moles HCl = volume (L) x concentration (M)
= 4.00 mL x (1 L/1000 mL) x 0.100 M
= 0.00400 moles
The reaction between HCl and NH3 can be represented by the following equation:
HCl + NH3 --> NH4+ + Cl-
Since HCl is a strong acid, it completely dissociates in water, so the reaction shifts to the right. As a result, the concentration of NH4+ in the buffer solution increases, while the NH3 concentration decreases.
To calculate the new concentrations, we need to consider the following:
[HCl] = [NH4+] = initial concentration + moles of HCl added / total volume
[NH3] = initial concentration - moles of HCl added / total volume
[HCl] = [NH4+] = 0.100 M + 0.00400 moles / (100.0 mL + 4.00 mL) = 0.104 M
[NH3] = 0.100 M - 0.00400 moles / (100.0 mL + 4.00 mL) = 0.0960 M
Finally, let's calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
= 9.25 + log (0.0960/0.104)
= 9.25 + log (0.923)
≈ 9.25 - 0.033
≈ 9.22
Therefore, the change in pH when 4.00 mL of 0.100 M HCl is added to 100.0 mL of the buffer solution is approximately -0.03 units (decrease in pH).
b) To calculate the change in pH when a base is added to a buffer solution, we follow similar steps as in part (a) but with a few modifications.
Since NaOH is a strong base, it completely dissociates in water, resulting in an increase in OH- concentration in the buffer solution. This increase in OH- concentration will react with NH4+ to form NH3 and H2O according to the following equation:
NH4+ + OH- --> NH3 + H2O
To calculate the final pH, we need to recalculate the concentrations of NH4+ and NH3 after the addition of NaOH.
First, let's calculate the moles of NaOH added:
moles NaOH = volume (L) x concentration (M)
= 4.00 mL x (1 L/1000 mL) x 0.100 M
= 0.00400 moles
Next, we determine the new concentrations of NH4+ and NH3:
[NH4+] = initial concentration - moles NaOH added / total volume
[NH3] = initial concentration + moles NaOH added / total volume
[NH4+] = 0.100 M - 0.00400 moles / (100.0 mL + 4.00 mL) = 0.0960 M
[NH3] = 0.100 M + 0.00400 moles / (100.0 mL + 4.00 mL) = 0.104 M
Now, let's calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
= 9.25 + log (0.104/0.0960)
= 9.25 + log (1.083)
≈ 9.25 + 0.036
≈ 9.29
Therefore, the change in pH when 4.00 mL of 0.100 M NaOH is added to the original buffer solution is approximately +0.04 units (increase in pH).