Given that 60.0 mL of 0.550 M NH4+ is titrated with 0.550 M NaOH. The Ka for NH4+ is 5.60x10^-10. Calculate the pH of the solution after 40.0 mL of NaOH has been added to the solution.
millimoles NH4^+ initially = mL x M = 60 x 0.550 = 33
millimoles NaOH added = 40 x 0.550 = 22
...............NH4^+ + NaOH ==> NH3 + H2O + Na^+
Initial........33..............0..............0.........0...........0
add.............................22.......................................
change...-22..............-22...........+22...................
equil..........11...............0..............22............................
Insert the equilibrium line into the Henderson-Hasslebalch equation.
pH = pKa + log [(base)/(acid)]
pH = + log (22/11) = ? or if you aren't familiar with the H-H equation, from the ionization of NH4^+ you get
...................NH4^+ ==> NH3 + H^+
Ka = (NH3)(H^+)/(NH4^+). Solve for
(H^+) = Ka(NH4+)/(NH3)
(H^+) = 5.6E-10*(11/22)
Then convert to pH.
Post your work if you get stuck.
Okay, thanks. That helped me a lot. How would you go about the same problem after 60.0mL of NaOH has been added to the solution? I think I know how to do it but I am not sure how to set up the equilibrium equation
To find the pH of the solution after adding 40.0 mL of NaOH, we need to determine the concentration of NH4+ and OH- after the reaction.
Step 1: Calculate the moles of NH4+ initially present:
Moles of NH4+ = Volume of NH4+ solution (L) x Concentration of NH4+ (mol/L)
Moles of NH4+ = (60.0 mL / 1000) L x 0.550 mol/L
Step 2: Calculate the moles of OH- added:
Moles of OH- = Volume of NaOH solution added (L) x Concentration of NaOH (mol/L)
Moles of OH- = (40.0 mL / 1000) L x 0.550 mol/L
Step 3: Calculate the remaining moles of NH4+:
Remaining moles of NH4+ = Initial moles of NH4+ - Moles of OH- added
Step 4: Calculate the remaining concentration of NH4+:
Remaining concentration of NH4+ = Remaining moles of NH4+ / Total volume of the solution (L)
Step 5: Calculate the concentration of NH3 (base) formed:
Concentration of NH3 = Remaining concentration of NH4+
Step 6: Calculate the concentration of OH-:
Concentration of OH- = Moles of OH- added / Total volume of the solution (L)
Step 7: Use the Kb to calculate the concentration of NH4OH (base):
Kb = Kw / Ka
Kb = 1.0 x 10^-14 / 5.60 x 10^-10
Step 8: Calculate the concentration of H+ (acid):
[H+] = Kw / [OH-]
[H+] = 1.0 x 10^-14 / Concentration of OH-
Step 9: Calculate the pH of the solution:
pH = -log[H+]
Now, let's plug in the values and calculate the pH.
Please note that Kb (base dissociation constant) = Kw (water dissociation constant) / Ka (acid dissociation constant).
To calculate the pH of the solution after 40.0 mL of NaOH has been added, we need to determine the moles of NH4+ and OH- remaining in the solution and use them to calculate the concentration of H+ ions. From there, we can use the equilibrium constant Ka for NH4+ to calculate the pH.
Here are the steps to follow:
1. Convert the volume of NH4+ solution to moles:
- 60.0 mL of NH4+ solution * (0.550 mol NH4+/L) = 0.550 * 60.0 / 1000 = 0.0330 mol NH4+
2. Calculate the moles of OH- added from NaOH:
- 40.0 mL of NaOH solution * (0.550 mol NaOH/L) = 0.550 * 40.0 / 1000 = 0.0220 mol OH-
3. Calculate the moles of NH4+ remaining:
- Initial moles of NH4+ - moles of OH- added = 0.0330 mol - 0.0220 mol = 0.0110 mol NH4+
4. Calculate the concentration of NH4+ in the remaining solution:
- Concentration = moles / volume = 0.0110 mol / (60.0 mL + 40.0 mL) = 0.0110 mol / 100 mL = 0.110 M NH4+
5. Calculate the concentration of NH3 (conjugate base of NH4+):
- NH3 is formed by the reaction: NH4+ + OH- → NH3 + H2O
- The moles of NH3 formed is equal to the moles of OH- added since the reaction is 1:1 in terms of moles.
- Concentration of NH3 = moles of NH3 / volume = 0.0220 mol / 100 mL = 0.220 M NH3
6. Calculate the concentration of H+ ions:
- Using the equilibrium constant Ka for NH4+ (Ka = [NH3][H+]/[NH4+]), we can set up the equation:
- Ka = [0.220][H+]/[0.110]
- Rearranging and solving for [H+]: [H+] = (Ka * [NH4+]) / [NH3]
7. Substitute the given values into the equation to find [H+]:
- [H+] = (5.60x10^-10 * 0.110) / 0.220
8. Calculate the pH:
- pH = -log[H+]
Following these steps, you can calculate the pH of the solution after 40.0 mL of NaOH has been added.