A 75.0 mL volume of 0.200 M NH3 (Kb=1.8x10^-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0 mL of HNO3.

To calculate the pH after the addition of 28.0 mL of HNO3, we need to consider the reaction between NH3 (ammonia) and HNO3 (nitric acid). Ammonia is a weak base, and nitric acid is a strong acid. The reaction can be represented as:

NH3 + HNO3 -> NH4+ + NO3-

Initially, we have a 75.0 mL volume of 0.200 M NH3. This means that there are 0.200 moles of NH3 in every liter of solution. To determine the number of moles of NH3 in 75.0 mL, we use the following conversion:

moles of NH3 = (0.200 M) x (0.075 L) = 0.015 moles

With the addition of 28.0 mL of 0.500 M HNO3, we can calculate the number of moles of HNO3 added:

moles of HNO3 = (0.500 M) x (0.028 L) = 0.014 moles

Now, let's calculate the number of moles of NH3 and HNO3 remaining after the reaction:

moles of NH3 remaining = initial moles of NH3 - moles of HNO3 added
= 0.015 moles - 0.014 moles
= 0.001 moles

Since NH3 is a weak base, it will react with HNO3 to form NH4+ (ammonium ion). This means that the concentration of NH4+ will increase. However, NH4+ is a weak acid and hydrolyzes in water. The hydrolysis reaction is:

NH4+ + H2O -> NH3 + H3O+

As we are interested in finding the pH, we need to calculate the concentration of H3O+ ions.

Since the initial volume of NH3 solution was 75.0 mL and we added an additional 28.0 mL of HNO3, the total volume of the solution is (75.0 + 28.0) mL = 103.0 mL.

To determine the new concentration of NH4+, we divide the number of moles of NH4+ by the new total volume of the solution:

NH4+ concentration = (moles of NH4+) / (total volume of solution in liters)
= (0.014 moles) / (0.103 L)
= 0.136 M

Since NH4+ hydrolyzes to produce H3O+ ions, the concentration of H3O+ will be equal to the concentration of NH4+.

Therefore, the concentration of H3O+ is 0.136 M.

To calculate the pH, we use the equation:

pH = -log[H3O+]
= -log(0.136)
= 0.866

Thus, after the addition of 28.0 mL of HNO3, the pH of the solution is approximately 0.866.

NH3 + HNO3 ==> NH4NO3

Set up an ICE chart, determine which reagent is in excess after the addition of the HNO3, the pH from that.