A 10.0-mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCl solution:
a) 0.00 mL
b) 10.0 mL
c) 20.0 mL
c) 30.0 mL
d) 40.0 mL
**I know how to solve the problem but I keep getting answers that are exactly one away from the correct answer.
Actually I figured out a, b, c, and d but I can't figure out e. What do you do if HCl is in excess?
If HCl is in excess, calculate how much excess in mols, divide by the TOTAL volume at that point to obtain (HCl) in mols/L; ie., M, then pH = -log(HCl) since HCl is a strong acid and ionizes 100%.
To calculate the pH after each addition of the HCl solution, we need to consider the reaction that occurs between NH3 (ammonia) and HCl (hydrochloric acid). Ammonia acts as a base, and hydrochloric acid acts as an acid.
The reaction between NH3 and HCl can be written as follows:
NH3 (aq) + HCl (aq) → NH4+ (aq) + Cl- (aq)
At the start of the titration, before adding any HCl solution (0.00 mL), we have a solution containing only NH3. Since NH3 is a weak base, it undergoes partial ionization and produces OH- ions in solution. The concentration of OH- ions can be calculated using the Kb value for NH3, which is 1.8 x 10^-5.
To calculate the concentration of OH- ions, we can set up an equilibrium expression:
Kb = [NH4+][OH-] / [NH3]
Since we have excess NH3 and no NH4+ at the start, we can assume that [NH4+] is negligibly small. Therefore, the equation simplifies to:
Kb = [OH-] * [NH3]
We can rearrange the equation to solve for [OH-]:
[OH-] = Kb / [NH3]
Using the given concentration of NH3 (0.300 M), we can calculate [OH-]:
[OH-] = (1.8 x 10^-5) / (0.300) = 6.0 x 10^-5 M
To find the pOH, we can take the negative logarithm (base 10) of [OH-]:
pOH = -log([OH-]) = -log(6.0 x 10^-5) ≈ 4.221
Finally, we can calculate the pH using the relationship:
pH = 14 - pOH = 14 - 4.221 ≈ 9.779
For each subsequent addition of the HCl solution, we need to consider the reaction stoichiometry and calculate the concentration of OH- ions based on the remaining NH3 and added HCl.
Repeat the calculations for each addition of HCl solution at 10.0 mL, 20.0 mL, 30.0 mL, and 40.0 mL to determine the corresponding pH values.
If you are consistently getting answers that are one unit away from the correct answer, double-check your calculations to ensure you haven't made any errors in decimal placement, unit conversions, or taking logarithms. Additionally, pay attention to any rounding errors that can accumulate and affect the final result.