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 determine the number of moles of HNO3 that reacted with NH3, and then use this information to calculate the concentration of the remaining NH3 and the resulting pH.
Let's start by finding the number of moles of NH3 initially present in the solution:
Molarity (M) = moles (mol) / volume(L)
0.200 M = moles NH3 / 0.0750 L
moles NH3 = 0.200 M * 0.0750 L = 0.015 mol
Next, we need to determine the number of moles of HNO3 that reacted with NH3. Since the reaction between NH3 and HNO3 occurs in a 1:1 ratio, the moles of HNO3 that reacted will be equal to the moles of NH3 initially present:
moles HNO3 reacting = 0.015 mol
Since we know the concentration of HNO3 is 0.500 M and the volume of HNO3 added is 28.0 mL, we can calculate the moles of HNO3 added:
moles HNO3 added = 0.500 M * 0.0280 L = 0.014 mol
Now we can calculate the moles of NH3 remaining after the reaction:
moles NH3 remaining = moles NH3 initially present - moles HNO3 reacting
moles NH3 remaining = 0.015 mol - 0.015 mol = 0 mol
After the reaction, all of the NH3 has been consumed, resulting in an excess of HNO3.
To find the concentration of NH3 remaining, we divide the moles of NH3 remaining by the total volume of the solution:
Concentration NH3 remaining (M) = moles NH3 remaining / total volume(L)
Concentration NH3 remaining = 0 mol / (0.0750 L + 0.0280 L) = 0 M
Since the concentration of NH3 remaining is now zero, we can consider the solution as pure HNO3, which is a strong acid. The pH of a strong acid is calculated using the equation:
pH = -log[H+]
To calculate the pH, we need to determine the concentration of H+ ions, which is equal to the concentration of HNO3:
[H+] = concentration HNO3 = 0.500 M
pH = -log(0.500) = 0.301
Therefore, the pH after the addition of 28.0 mL of HNO3 is approximately 0.301.