Question

A. A 75.0-mL volume of 0.200 mol L^-1 NH3 (Kb=1.8×10^−5) is titrated with 0.500 mol L^−1 HNO3. Calculate the pH after the addition of 13.0 mL of HNO3 .

B. A 52.0-mL volume of 0.350 mol L^−1 CH3COOH (Ka=1.8×10^−5. ) is titrated with 0.400 mol L^−1 NaOH. Calculate the pH after the addition of 19.0 mL of NaOH.

Answer 1

To calculate the pH after the addition of 13.0 mL of HNO3, we can use stoichiometric calculations to determine the remaining concentration of NH3. From there, we can calculate the pOH and pH of the solution.

Step-by-step explanation:

To calculate the pH after the addition of 13.0 mL of HNO3, we need to consider the reaction between NH3 and HNO3. NH3 is a weak base, and HNO3 is a strong acid.

First, calculate the number of moles of NH3 present in the initial solution: 0.0750 L x 0.200 mol/L = 0.015 mol.

Next, calculate the number of moles of HNO3 added: 0.013 L x 0.500 mol/L = 0.0065 mol.

Using the reaction equation NH3 + HNO3 → NH4NO3, we can see that 0.0065 mol of HNO3 will react with 0.0065 mol of NH3. This leaves 0.015 - 0.0065 = 0.0085 mol of NH3 in the solution.

To calculate the concentration of NH3 in the solution, divide the remaining moles of NH3 by the new volume of the solution: 0.0085 mol / (0.0750 L + 0.013 L) = 0.093 mol/L.

Finally, calculate the pOH using the equation pOH = -log[OH-] and then calculate the pH using the equation pH = 14 - pOH. The pOH is -log[10-14] = 14, and therefore the pH is 14 - 14 = 0. The solution is acidic.