A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3 with 50 mL of 0.300 NH4Cl. The pKb of NH3 is 4.74.
NH3 + H2O-> NH4+ +OH-
7.50 mL of 0.125 M HCl is added to the 100 mL of the buffer solution. Calculate the concentration of NH3 and NH4Cl for the buffer solution. Calculate the pH of the solution.
Okay thank you so much!
Why do I subtract .9375 from H^+ instead of adding .9375 and then subtracting .9375 from NH4^+ instead of adding?
With regard to H^+. You added 0.9375 mols H^+ to the buffer. ALL of it will react with the 15.0 mmols NH3 (H^+ is the limiting reagent so there won't be any of it left). With regard to the NH4+ you should add and my ICE chart shows that; however, I said subtract in my instructions. I should have said, "Add the I line to the C line algebraically) to arrive at the E line. That might have cleared up the H^+ question too if I had stated my instructions correctly.
I have one question. When you say divide by the total mL, did you include the 7.5mL of HCl that you added?
To calculate the concentration of NH3 and NH4Cl in the buffer solution after adding 7.50 mL of 0.125 M HCl, we need to follow these steps:
1. Determine the initial moles of NH3 and NH4Cl before adding HCl:
- Moles of NH3 = volume (L) x concentration (M) = 0.050 L x 0.300 M
- Moles of NH4Cl = volume (L) x concentration (M) = 0.050 L x 0.300 M
2. Calculate the moles of HCl added:
- Moles of HCl = volume (L) x concentration (M) = 0.0075 L x 0.125 M
3. Determine the net change in moles of NH3 and NH4Cl after the reaction with HCl:
- NH3 reacts with HCl to form NH4+:
- The stoichiometric ratio is 1:1, so the moles of NH3 decrease by the same amount as the moles of HCl added.
- NH4Cl does not react with HCl, so its moles remain the same.
4. Calculate the final moles of NH3 and NH4Cl after the reaction:
- Final moles of NH3 = initial moles of NH3 - moles of HCl added
- Final moles of NH4Cl = initial moles of NH4Cl
5. Calculate the final concentration of NH3 and NH4Cl:
- Final concentration of NH3 = final moles of NH3 / total volume (L)
- Final concentration of NH4Cl = final moles of NH4Cl / total volume (L)
Now, let's plug in the values:
1. Initial moles of NH3:
- Moles of NH3 = 0.050 L x 0.300 M = 0.015 mol
2. Initial moles of NH4Cl:
- Moles of NH4Cl = 0.050 L x 0.300 M = 0.015 mol
3. Moles of HCl added:
- Moles of HCl = 0.0075 L x 0.125 M = 0.0009375 mol
4. Final moles of NH3:
- Final moles of NH3 = 0.015 mol - 0.0009375 mol = 0.0140625 mol
5. Final moles of NH4Cl:
- Final moles of NH4Cl = 0.015 mol
6. Total volume of the solution:
- Total volume = initial volume of NH3 + initial volume of NH4Cl + volume of HCl added
- Total volume = 0.050 L + 0.050 L + 0.0075 L = 0.1075 L
7. Final concentration of NH3:
- Final concentration of NH3 = 0.0140625 mol / 0.1075 L = 0.1307 M
8. Final concentration of NH4Cl:
- Final concentration of NH4Cl = 0.015 mol / 0.1075 L = 0.1395 M
To calculate the pH of the solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
The pKa can be calculated from the pKb of NH3:
pKa = 14 - pKb
Now, let's calculate the pH:
pKa = 14 - 4.74 = 9.26
[A-] = concentration of NH4Cl = 0.1395 M
[HA] = concentration of NH3 = 0.1307 M
pH = 9.26 + log (0.1395 / 0.1307)
pH = 9.26 + log (1.064)
pH = 9.26 + 0.027
pH = 9.29
Therefore, the concentration of NH3 is 0.1307 M and the concentration of NH4Cl is 0.1395 M in the buffer solution after adding 7.50 mL of 0.125 M HCl. The pH of the solution is 9.29.
millimols NH3 = 50 mL x 0.3M = 15.0
(NH3) = mmols/mL = 15.0/100 = 0.15M initially.
millimols NH4Cl = 50 x 0.3 = 15.0
(NH4Cl) = 15.0/100 = 0.15M initially
millimols HCl = 7.50 x 0.125 = 0.9375
pKb NH3 = 4.74; pKa = 14-4.74 = 9.26
I prefer to work with this equilibrium in mmols.
...........NH3 + H^+ ==> NH4^+
I..........15....0........15
add............0.9375..............
C......-0.9375..-0.9375...+0.9375
E.........???....???.......????
Subtraqct I-C = E line and substitute those values into HH equation solve for pH.
Final (NH3), (NH4Cl) = mmols each/total mL.
Post your work if you get stuck.