Calculate the pH of 100.0 mL ofa buffer that is 0.070 M NH4CL and 0.130 M NH3 before and after the addition of 1.00 ml of 5.55 M HNO3.
Before = ?
After = ?
I know the base is NH3 and the acid is NH4Cl!
millimols NH3 = M x mL = ?
mmols NH4Cl = mL x M = ?
mmols HNO3 = mL x M = ?
Before:
Use the Henderson-Hasselalch equation.
After:
Construct an ICE chart frp, below and add the HNO3. Recalculate using the HH equation.
............NH3 + H^+ ==> HNH4^+
initial.....
add HNO3 ........
change
equil.
To calculate the pH of the buffer solution before and after the addition of the 1.00 mL of 5.55 M HNO3, you need to consider the Henderson-Hasselbalch equation for the buffer solution.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
Where:
pH is the measure of acidity or alkalinity
pKa is the logarithmic acid dissociation constant of the weak acid component of the buffer
[A-] is the concentration of the conjugate base
[HA] is the concentration of the weak acid
1. Calculate the initial pH before the addition of the HNO3:
Given:
Volume of buffer solution = 100.0 mL = 0.100 L
Concentration of NH4Cl = 0.070 M
Concentration of NH3 = 0.130 M
First, you need to find the pKa value for the NH4Cl-NH3 buffer system. The pKa value can be obtained from reference sources or calculated using the acid dissociation constant (Ka) of the acid (NH4Cl).
Next, you need to calculate the concentrations of the NH4+ and NH3 ions:
[HA] = [NH4+] = 0.070 M
[A-] = [NH3] = 0.130 M
Now, substitute the given values into the Henderson-Hasselbalch equation:
pH = pKa + log ([NH3] / [NH4+])
Solve for pH by substituting the values:
pH = pKa + log (0.130 / 0.070)
2. Calculate the pH after the addition of 1.00 mL of 5.55 M HNO3:
To calculate the pH after the addition of the HNO3, you need to consider the reaction between the HNO3 and NH3 to form NH4+.
Given:
Volume of HNO3 added = 1.00 mL = 0.001 L
Concentration of HNO3 = 5.55 M
To determine how much of the NH3 reacts with the HNO3, we need to calculate the moles of HNO3 added:
Moles of HNO3 = concentration x volume
= 5.55 M x 0.001 L
From the stoichiometry of the reaction between HNO3 and NH3, we know that one mole of HNO3 reacts with one mole of NH3 to form one mole of NH4+.
Therefore, the moles of HNO3 reacted are equal to the moles of NH3 consumed and the moles of NH4+ formed.
Now, calculate the new concentrations of NH4+ and NH3 after the reaction:
[HA] = [NH4+] = (initial concentration of NH4+) - moles of NH4+ formed
= 0.070 M - (moles of HNO3 reacted / total volume of buffer solution)
[A-] = [NH3] = (initial concentration of NH3) - moles of NH3 consumed
= 0.130 M - moles of HNO3 reacted / total volume of buffer solution
Finally, substitute the new concentrations into the Henderson-Hasselbalch equation:
pH = pKa + log ([NH3] / [NH4+])
Solve for pH by substituting the new values:
pH = pKa + log ([NH3] / [NH4+])