Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8*10^-5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant.
A 30.0-ml volume of 0.50 M CH3COOH (Ka=1.8*10^-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH.
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Both of these problems have moles acid = moles base which means you are at the equivalence point of each. Therefore, the pH is determined by the pH of the salt.
Write the hydrolysis equation, set up an ICE chart, and solve for (H^+), then convert to pH. One important point to remember (and easy to forget) is that the concn of the salt form is moles acid or base/TOTAL volume at the equivalence point.
To calculate the pH after the addition of 50.0 mL of the titrant, we need to determine the amount of base and acid that react and calculate the resulting concentrations of the conjugate acid and base.
Let's break down the steps to solve the problem:
Step 1: Calculate the amount of NH3 (base) and HNO3 (acid) that react.
The volume of the base (NH3) is 50.0 mL, and its concentration is 0.20 M. Volume × concentration = amount in moles.
Amount of NH3 (base) = 50.0 mL × 0.20 M = 10.0 mmol
Since HNO3 is a strong acid, it will completely dissociate, so the amount of HNO3 (acid) is also 10.0 mmol.
Step 2: Determine the limiting reactant.
To determine which reactant is the limiting reactant, we need to compare the amounts of NH3 and HNO3. Since both have the same amount (10.0 mmol), neither is in excess. Therefore, both reactants will be consumed completely.
Step 3: Calculate the resulting concentrations.
The reaction between NH3 and HNO3 produces NH4+ and NO3-. NH4+ is the conjugate acid of NH3, while NO3- is a spectator ion.
Given that the initial volume is 50.0 mL and the total volume after addition is 100.0 mL (50.0 mL base + 50.0 mL acid), we can calculate the final concentration of NH4+:
Final volume = 100.0 mL = 0.100 L (converting mL to L)
Final concentration of NH4+ = 10.0 mmol / 0.100 L = 0.10 M
Step 4: Calculate the pH.
To calculate the pH, we need to use the pKa value for the reaction involving NH4+ and H2O. Since NH4+ is the conjugate acid of NH3, we need to find the pKa value for NH3/NH4+ system.
pKa = -log10(Ka)
Given that Kb (dissociation constant of NH3) is 1.8 × 10^-5, we can use the relationship:
Kw = Ka × Kb
Kw (ion product of water) = 1.0 × 10^-14 at 25°C
Substituting the values:
1.0 × 10^-14 = 1.8 × 10^-5 × Ka
Rearranging the equation:
Ka = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10
Now, we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([NH4+] / [NH3])
pKa = -log(5.56 × 10^-10) ≈ 9.26
[NH4+] is the final concentration of NH4+ = 0.10 M
[NH3] is the initial concentration of NH3 = 0.20 M
pH = 9.26 + log(0.10 / 0.20)
pH = 9.26 + log(0.5)
pH = 9.26 - 0.30
pH ≈ 8.96
Therefore, the pH after the addition of 50.0 mL of the titrant is approximately 8.96.