35.0mL of 0.050 M HCl is titrated with 0.050 M NH3 (Kb= 1.8 x 10^-5). The pH of the analyte solution after 15.0mL addition of NH3 titrant is?

NH3 + HCl ==> NH4Cl

mols NH3 = M x L = ?
mols HCl = M x L = ?
Subtract mols HCl - mols NH3. I think HCl is in excess but you should confirm that. .
That means pH = =log (HCl).

To find the pH of the analyte solution after adding 15.0 mL of NH3 titrant, we need to consider the reaction between HCl and NH3 and determine the concentration of the resulting species.

First, let's determine the amount of HCl present in the analyte solution before the addition of NH3 titrant. We have 35.0 mL of 0.050 M HCl:

Amount of HCl = Volume × Concentration
= 35.0 mL × 0.050 M
= 1.75 mmol

Next, let's determine the amount of NH3 titrant added. We have added 15.0 mL of 0.050 M NH3:

Amount of NH3 = Volume × Concentration
= 15.0 mL × 0.050 M
= 0.75 mmol

The reaction between HCl and NH3 can be represented as follows:

HCl + NH3 → NH4+ + Cl-

Since HCl is a strong acid, it will completely dissociate, while NH3 is a weak base. We can assume that the reaction goes to completion and that all the NH3 is consumed in the reaction to form NH4+ and Cl- ions.

Now, let's calculate the concentration of NH4+ that forms. The amount of NH4+ formed will be equal to the amount of NH3 initially present:

Amount of NH4+ = Amount of NH3
= 0.75 mmol

To find the concentration of NH4+ in the solution, we need to take into account the total volume of the solution after the NH3 addition. Since we added 15.0 mL of NH3 titrant to the initial 35.0 mL HCl solution, the total volume becomes 50.0 mL (35.0 mL + 15.0 mL).

Concentration of NH4+ = Amount of NH4+ / Total Volume
= 0.75 mmol / 50.0 mL = 0.015 M

To calculate the pOH of the solution, we need to use the equilibrium expression for the reaction of NH4+ with water. NH4+ is the conjugate acid of NH3 and can donate a proton:

NH4+ + H2O ⇌ NH3 + H3O+

The Kb value of NH3 is given as 1.8 x 10^-5, which is the equilibrium constant expression for the reaction. From the Kb expression, we can write:

Kb = [NH3][H3O+] / [NH4+]

Since we initially have negligible [H3O+], we can assume it to be zero. Therefore, the expression simplifies to:

Kb = [NH3] / [NH4+]

Plugging in the values:

1.8 x 10^-5 = 0.050 M / [NH4+]
[NH4+] = 0.050 M / 1.8 x 10^-5
[NH4+] ≈ 2.78 M

Finally, we can calculate the pOH:

pOH = -log10 [NH4+]
= -log10 (2.78)
≈ 0.556

The pH of the analyte solution can then be found using the relation:

pH + pOH = 14

pH = 14 - pOH
= 14 - 0.556
≈ 13.44

Therefore, the pH of the analyte solution after adding 15.0 mL of NH3 titrant is approximately 13.44.