Find the pH and volume (mL) of 0.447 M HNO3 needed to reach the equivalence point(s) in titrations of each of the following.
(a) 2.65 L of 0.0950 M pyridine (C5H5N)
pH
volume
Answer = 563 mL
(b) 0.188 L of 0.250 M NH3
pH
volume Answer = 105 mL
I can't figure out the pHs for either
never mind, I figured this one out. Thanks!
To find the pH at the equivalence point in each titration, we need to calculate the concentrations of the resulting solutions after the reaction between the acid and base has completed.
(a) Titrating 2.65 L of 0.0950 M pyridine (C5H5N) with 0.447 M HNO3:
1. Write the balanced chemical equation for the reaction between pyridine (C5H5N) and nitric acid (HNO3):
C5H5N + HNO3 -> C5H5NH+ + NO3-
2. Determine the number of moles of each species in the reaction:
- Pyridine (C5H5N):
Moles = Volume (L) * Concentration (M)
Moles = 2.65 L * 0.0950 M = 0.25175 mol
- Nitric Acid (HNO3):
Moles = Volume (L) * Concentration (M)
Moles = ? * 0.447 M
Since we don't know the volume of the HNO3 solution required to reach the equivalence point, we proceed further.
3. At the equivalence point, the moles of HNO3 reacting with pyridine will be equal to the moles of pyridine present initially. Therefore:
Moles of HNO3 = Moles of C5H5N
0.25175 mol = ? * 0.447 M
Solving for the volume of HNO3 needed:
Volume (L) = Moles / Concentration
Volume = 0.25175 mol / 0.447 M ≈ 0.563 L ≈ 563 mL
So, the volume of 0.447 M HNO3 needed to reach the equivalence point in this titration is approximately 563 mL.
To calculate the pH, you would need additional information about pyridine as a base and its acid dissociation constant (Ka).
(b) Titrating 0.188 L of 0.250 M NH3 with 0.447 M HNO3:
1. Write the balanced chemical equation for the reaction between NH3 and HNO3:
NH3 + HNO3 -> NH4+ + NO3-
2. Determine the number of moles of each species in the reaction:
- NH3:
Moles = Volume (L) * Concentration (M)
Moles = 0.188 L * 0.250 M = 0.047 mol
- Nitric Acid (HNO3):
Moles = Volume (L) * Concentration (M)
Moles = ? * 0.447 M
Since we don't know the volume of the HNO3 solution required to reach the equivalence point, we proceed further.
3. At the equivalence point, the moles of HNO3 reacting with NH3 will be equal to the moles of NH3 present initially. Therefore:
Moles of HNO3 = Moles of NH3
0.047 mol = ? * 0.447 M
Solving for the volume of HNO3 needed:
Volume (L) = Moles / Concentration
Volume = 0.047 mol / 0.447 M ≈ 0.105 L ≈ 105 mL
So, the volume of 0.447 M HNO3 needed to reach the equivalence point in this titration is approximately 105 mL.
Similar to the previous question, to calculate the pH, additional information about NH3 as a base and its acid dissociation constant (Kb) is required.
To determine the pH at the equivalence point in a titration, you need to identify the nature of the acid and the base involved. In these titrations, HNO3 is the acid and pyridine (C5H5N) or NH3 are bases.
(a) Titrating HNO3 with pyridine:
1. Determine the moles of pyridine:
moles of pyridine = concentration of pyridine (mol/L) × volume of pyridine (L)
moles of pyridine = 0.0950 M × 2.65 L
2. Since pyridine is a base, it will react with the acid, HNO3, in a 1:1 ratio. This means the moles of HNO3 required will be equal to the moles of pyridine.
moles of HNO3 = moles of pyridine
3. Calculate the volume of 0.447 M HNO3 required to reach the equivalence point:
volume of HNO3 (L) = moles of HNO3 / concentration of HNO3 (mol/L)
volume of HNO3 = moles of pyridine / 0.447 M
4. Convert the volume to mL:
volume of HNO3 (mL) = volume of HNO3 (L) × 1000
Substitute the values to calculate the pH.
(b) Titrating HNO3 with NH3:
Perform the same steps as in part (a) by substituting the values for NH3. Again, since NH3 is a base, it will react with HNO3 in a 1:1 ratio.
Once you have the volume of HNO3 required to reach the equivalence point, you can use this value to calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Note: You will need to know the pKa of the acid (HNO3) and the concentrations of the acid and conjugate base (A-) to calculate the pH.