Consider the titration of 50.0 mL of 0.20 M NH3 (Kb = 1.8x10^-5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.
5.13
To calculate the pH after addition of 50.0 mL of the titrant in the titration of NH3 with HNO3, we need to determine the concentration of the excess HNO3 and NH4+ ions formed as a result of the reaction.
Here's how you can approach this problem:
1. Write the balanced equation for the reaction between NH3 and HNO3:
NH3 + HNO3 → NH4+ + NO3-
2. Determine the initial moles of NH3 in the solution:
Moles of NH3 = Initial concentration of NH3 x Volume of NH3 solution
= 0.20 M x 0.050 L
= 0.0100 moles
3. Since NH3 and HNO3 react in a 1:1 ratio, the moles of HNO3 added will be equal to the moles of NH3:
Moles of HNO3 = 0.0100 moles
4. Calculate the concentration of HNO3 after the addition of 50.0 mL:
Concentration of HNO3 = Moles of HNO3 / Volume of HNO3 solution
= 0.0100 moles / 0.050 L
= 0.20 M (unchanged)
5. Calculate the concentration of NH4+ ions formed:
Concentration of NH4+ = Moles of NH4+ ions / Total volume of the solution
= 0.0100 moles / 0.100 L (50 mL NH3 + 50 mL HNO3)
= 0.10 M
6. Use the relation between Kb (base dissociation constant) and the concentration of NH4+ and NH3 to find the concentration of NH3:
Kb = [NH4+][OH-] / [NH3]
1.8x10^-5 = (0.10 M)([OH-]) / (0.10 M)
[OH-] = 1.8x10^-5 M
7. Calculate the concentration of H3O+ ions (from the self-ionization of water) using the relation: [H3O+][OH-] = 1.0x10^-14:
[H3O+] = 1.0x10^-14 / (1.8x10^-5 M)
= 5.6x10^-10 M
8. Finally, calculate the pH using the relation: pH = -log[H3O+]:
pH = -log(5.6x10^-10)
≈ 9.25
Therefore, the pH after the addition of 50.0 mL of the titrant is approximately 9.25.
8.5
The secret to doing titration problems is to recognize what you have in the solution. In this case, the NH3 is exactly neutralized by the HNO3; therefore, you are at the equivalence point and you have NH4NO3 is solution. How much NH4NO3? That will be 50.0 mL x 0.20 M = 10 millimoles in 100 mL = 0.1 M
Write the hydrolysis equation.
NH4^+ + HOH ==> NH3 + H3O^+
Then do the ICE bit.
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+
Solve for (H3O^+) and convert to pH.