A 1M solution (200ml) of NH3 (Kb of ammonia=1.8*10-5) is added to 200ml of 0.5M HCl. Calculate:
a) the resulting pH?
b) The delta ph due to addition of 15ml of 1M HNO3 to solution obtained in part a
c) I am stuck on this:
The delta ph due to addition of 15ml of 1M NaOH to the solution obtained in part a
NH3 + HCl --> NH4Cl + H2O
millimols NH3 = 200 x 1 = 200
mmols HCl added = 200 x 0.5 = 100
So you have 100 mmols NH3 in 400 mL solution. I assume from your post that you worked the remainder of part a.
For part b we have the soln from part a and add 15 mmols (15x1) NaOH.
So the OH^- in the soln consists of the OH^- from the ionization of NH3 plus the OH^- from the NaOH. You should note that the OH^- from NaOH will suppress the ionization of NH3.
The OH^- from NaOH = 15 mmols/415 mL = approx 0.0361 but you should confirm that and work it more accurately if needed.
To this you need to add the OH^- from the NH3. I suspect this is negligible (but not zero) but you must go through and see.
NH3 + H2O ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
(NH4^+) = x
(OH^-) = x + 0.0361 and you assume x is small so OH^- is 0.0361M
(NH3) = 100/415 = about 0.241M
Solve for x = about 1E-4 which shows x is small in comparison to the OH^- from NaOH. Calculate pH from this and then delta pH.
To calculate the resulting pH in part a, we first need to calculate the moles of NH3 and HCl present in the solution:
Moles of NH3 = Molarity * Volume = 1M * 0.2L = 0.2 moles
Moles of HCl = Molarity * Volume = 0.5M * 0.2L = 0.1 moles
Now, since NH3 is a base and HCl is an acid, they will react to form NH4+ and Cl- ions. The amount of NH3 that reacts with HCl is equal to the amount of HCl initially present. Therefore, after the reaction, we will be left with 0.1 moles of NH3 and NH4+ ions.
Now we can calculate the concentration of NH3 and NH4+ ions in the final solution:
Concentration of NH3/NH4+ = Moles / Total Volume = 0.1 moles / (0.2L + 0.2L) = 0.1 M
Since NH4+ is a weak acid, we can calculate its concentration using the dissociation constant Kb.
Kb = [NH4+][OH-] / [NH3]
0.1 x [OH-] / 0.1 = 1.8 x 10^-5
[OH-] = (1.8 x 10^-5) / 0.1
[OH-] = 1.8 x 10^-4
Now, we can calculate the pOH:
pOH = -log10([OH-])
pOH = -log10(1.8 x 10^-4)
pOH = 3.74
Finally, we can calculate the pH:
pH = 14 - pOH
pH = 14 - 3.74
pH = 10.26
a) The resulting pH is approximately 10.26.
Now moving to part b, we are adding 15 ml of 1M HNO3 to the solution obtained in part a.
Since HNO3 is a strong acid, it will completely dissociate in the water to form H+ ions. Therefore, the concentration of H+ ions will be equal to the concentration of HNO3, which is 1M.
Now, since the concentration of H+ ions is changing from 0.1M to 1M, we can calculate the change in pH.
Change in pH = -log10(new H+ concentration) - (-log10(initial H+ concentration))
Change in pH = -log10(1) - (-log10(0.1))
Change in pH = 0 - (-1)
Change in pH = 1
b) The delta pH due to the addition of 15ml of 1M HNO3 is 1.
Moving on to part c, we are adding 15 ml of 1M NaOH to the solution obtained in part a.
Since NaOH is a strong base, it will completely dissociate in the water to form Na+ and OH- ions. The concentration of Na+ ions is irrelevant to calculating the change in pH. We only need to consider the OH- ions.
The concentration of OH- ions added can be calculated from the concentration of NaOH, which is 1M. Since the volume being added is 15 ml, the moles of OH- ions added can be calculated as follows:
Moles of OH- = Molarity * Volume = 1M * 0.015L = 0.015 moles
Now, we need to calculate the new concentration of OH- ions in the solution:
Final concentration of OH- = (Initial concentration of OH- + moles of OH- added) / (Total volume)
Final concentration of OH- = (0.1M + 0.015 moles) / (0.2L + 0.015L)
Final concentration of OH- = (0.1M + 0.015) / (0.215L)
Final concentration of OH- = 0.523 M
Now, we can calculate the pOH:
pOH = -log10([OH-])
pOH = -log10(0.523)
pOH = 0.281
Finally, we can calculate the pH:
pH = 14 - pOH
pH = 14 - 0.281
pH = 13.719
c) The delta pH due to the addition of 15ml of 1M NaOH is approximately -13.72 (the negative sign indicates a decrease in pH).
To calculate the resulting pH, delta pH, and delta pH due to the addition of NaOH, we need to follow these steps:
a) Calculating the resulting pH after adding NH3 to HCl:
Step 1: Calculate the moles of NH3 and HCl:
NH3 moles = volume (200 ml) * concentration (1 M) = 0.2 mol
HCl moles = volume (200 ml) * concentration (0.5 M) = 0.1 mol
Step 2: Determine the limiting reactant:
Since NH3 and HCl react in a 1:1 ratio, the limiting reactant is the one with fewer moles, which, in this case, is HCl.
Step 3: Determine excess reactant:
NH3 is the excess reactant in this case.
Step 4: Calculate the moles of the excess reactant left after the reaction:
The moles of NH3 left is equal to the initial moles of NH3 minus the consumed moles of NH3 (from the reaction with HCl):
Moles of NH3 left = initial moles of NH3 - consumed moles of NH3
Step 5: Calculate the concentration of NH3 after the reaction:
Concentration of NH3 = moles of NH3 left / final volume of the solution
Final volume = initial volume of NH3 + initial volume of HCl (both 200 ml)
Step 6: Calculate pOH using the Kb value of NH3:
pOH = -log10(Kb) + log10(concentration of NH3)
Step 7: Calculate pH using the relation:
pH + pOH = 14, or pH = 14 - pOH
b) Calculating the delta pH due to the addition of 15 ml of 1M HNO3:
Following a similar methodology as in part a:
Step 1: Calculate the moles of HNO3:
HNO3 moles = volume (15 ml) * concentration (1 M) = 0.015 mol
Step 2: Determine the limiting reactant:
HNO3 is the limiting reactant in this case.
Step 3: Determine excess reactant:
NH3 is the excess reactant.
Step 4: Calculate the moles of the excess reactant left after the reaction:
The moles of NH3 left is equal to the initial moles of NH3 minus the consumed moles of NH3 (from the reaction with HNO3):
Moles of NH3 left = initial moles of NH3 - consumed moles of NH3
Step 5: Calculate the concentration of NH3 after the reaction:
Concentration of NH3 = moles of NH3 left / final volume of the solution
Final volume = initial volume of NH3 + initial volume of HCl + volume of HNO3 (all three are 200 ml)
Step 6: Calculate the new pOH using the Kb value of NH3:
New pOH = -log10(Kb) + log10(concentration of NH3)
Step 7: Calculate the new pH using the relation:
New pH = 14 - new pOH
c) Calculating the delta pH due to the addition of 15ml of 1M NaOH:
This step is similar to part b:
Step 1: Calculate the moles of NaOH:
NaOH moles = volume (15 ml) * concentration (1 M) = 0.015 mol
Step 2: Determine the limiting reactant:
NaOH is the limiting reactant in this case.
Step 3: Determine excess reactant:
HCl is the excess reactant.
Step 4: Calculate the moles of the excess reactant left after the reaction:
The moles of HCl left is equal to the initial moles of HCl minus the consumed moles of HCl (from the reaction with NaOH):
Moles of HCl left = initial moles of HCl - consumed moles of HCl
Step 5: Calculate the concentration of HCl after the reaction:
Concentration of HCl = moles of HCl left / final volume of the solution
Final volume = initial volume of NH3 + initial volume of HCl + volume of HNO3 + volume of NaOH (all four are 200 ml)
Step 6: Calculate the new pH using the relation:
New pH = -log10(concentration of HCl)