A 50 ml sample fo 0.240M NH3(aq) is titrated with aqueous hydrochloric acid. What is the pH after the addition fo 15.0 ml of 0.0600M HCl(aq)? (kb of NH3 = 1.8 x 10^5)

I believe this is a buffer solution problem so you should use the Henderson-Hasselbalch equation.

pH = pKa + log [(base)/(acid)]
NH3 + HCl ==> NH4Cl
How much NH3 do you have to start with?
M x L = ??
How much HCl has reacted with it.
M x L = ??
Subtract to see how much and what is in excess and substitute into the HH equation.
Remember that Ka = Kw/Kb

To find the pH after the addition of 15.0 ml of 0.0600 M HCl(aq) to a 50 ml sample of 0.240 M NH3(aq), we need to consider the following steps:

Step 1: Determine the moles of NH3 initially present.
To find the initial moles of NH3, we use the given concentration and volume:
moles of NH3 = concentration of NH3 * volume of NH3
moles of NH3 = 0.240 M * 50 mL
moles of NH3 = 0.240 mol/L * 0.050 L
moles of NH3 = 0.0120 mol

Step 2: Determine the moles of HCl added.
To find the moles of HCl added, we use the given concentration and volume:
moles of HCl = concentration of HCl * volume of HCl
moles of HCl = 0.0600 M * 15.0 mL
moles of HCl = 0.0600 mol/L * 0.015 L
moles of HCl = 0.000900 mol

Step 3: Determine the moles of NH3 remaining.
To find the moles of NH3 remaining after the reaction with HCl, we need to take into account the stoichiometry of the reaction. One mole of HCl reacts with one mole of NH3. Therefore, the moles of HCl added are equal to the moles of NH3 consumed.
moles of NH3 remaining = initial moles of NH3 - moles of HCl added
moles of NH3 remaining = 0.0120 mol - 0.000900 mol
moles of NH3 remaining = 0.0111 mol

Step 4: Calculate the concentration of NH3 after the reaction.
To find the concentration of NH3 after the reaction, we divide the moles of NH3 remaining by the total volume of the solution:
concentration of NH3 = moles of NH3 remaining / total volume of solution
total volume of solution = volume of NH3 + volume of HCl
total volume of solution = 50 mL + 15 mL = 65 mL = 0.065 L
concentration of NH3 = 0.0111 mol / 0.065 L
concentration of NH3 = 0.171 M

Step 5: Calculate the equilibrium constant (Kb) for the reaction.
The equilibrium constant (Kb) is given as 1.8 x 10^5. The Kb expression for the reaction is:
Kb = [NH4+][OH-] / [NH3]
Since NH3 is a weak base, it reacts with HCl to form NH4+ and Cl- ions. Therefore, NH4+ is the conjugate acid of NH3 and OH- is formed from the dissociation of water.
We can rearrange the Kb expression to solve for [OH-]:
[OH-] = (Kb * [NH3]) / [NH4+]

Step 6: Calculate the concentration of OH- and pH.
Using the calculated concentration of NH3 and Kb value, we can find the concentration of NH4+:
[NH4+] = (Kb * [NH3]) / [OH-]
Let's substitute the values into the equation:
[NH4+] = (1.8 x 10^5 * 0.171 M) / [OH-]
[NH4+] = 30780 / [OH-]

Since NH4+ and OH- combine to form water in a 1:1 ratio, the concentration of OH- is equal to the concentration of NH4+.

Therefore, [OH-] = [NH4+] = 0.171 M

Finally, we can calculate the pOH using the concentration of OH-:
pOH = -log([OH-])
pOH = -log(0.171)
pOH ≈ 0.766

To find the pH, we can use the equation:
pH + pOH = 14
pH + 0.766 = 14
pH ≈ 13.23

So, the pH after the addition of 15.0 ml of 0.0600 M HCl(aq) is approximately 13.23.