Calculate the change in pH when 5.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). (I have several problems like this can some one work this through for me as a base line)?
millimoles NH3 = mL x M = 100 x 0.1 = 10
mmols NH4^+ = 100 x 0.1 = 10
mmols HCl added = 5 x 0.1 = 0.5
.......NH3 + H^+ ==> NH4^+
I.....10.....0........10
added.......0.5...............
C....-0.5..-0.5......+0.5
E.....9.5.....0......+10.5
pH = pKa + log(base)/(acid)
pH = pKa + log(9.5/10.5)
pH = ?
Sure, I can help you work through this problem. To calculate the change in pH when an acid is added to a buffer solution, we need to consider the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the measure of acidity or alkalinity.
- pKa is the negative logarithm of the acid dissociation constant. It represents the strength of the acid.
- [A-] is the concentration of the conjugate base.
- [HA] is the concentration of the acid.
In this case, we have a buffer solution containing NH3 (conjugate base) and NH4Cl (acid). The pKa of NH4Cl can be obtained from tables as 9.24.
Now, let's calculate the pH of the buffer solution before the addition of HCl.
pH = pKa + log([A-]/[HA])
pH = 9.24 + log(0.100/0.100)
pH = 9.24 + log(1)
pH = 9.24 + 0
pH = 9.24
So, the pH of the buffer solution before the addition of HCl is 9.24.
To calculate the change in pH when 5.00 mL of 0.100 M HCl is added, we need to consider the stoichiometry and the resulting concentrations of NH3 and NH4Cl.
First, calculate the moles of HCl added:
moles HCl = volume x concentration
moles HCl = 5.00 mL (convert to L) x 0.100 M
moles HCl = 0.00500 L x 0.100 mol/L
moles HCl = 0.0005 mol
Since HCl is a strong acid, it will completely dissociate into H+ ions and Cl- ions.
The moles of NH3 will react with the added HCl to form NH4+ ions.
Since the volumes are given as 5.00 mL and 100.0 mL, we need to convert them to liters.
The resulting volume will be 105.0 mL, or 0.105 L.
Now let's calculate the new concentrations of NH3 and NH4Cl.
Concentration of NH3 = moles NH3 / volume in liters
Concentration of NH3 = (0.100 mol - 0.0005 mol) / 0.105 L
Concentration of NH3 = 0.00095 mol / 0.105 L
Concentration of NH3 = 0.00905 M
Concentration of NH4Cl = moles NH4Cl / volume in liters
Concentration of NH4Cl = 0.0005 mol / 0.105 L
Concentration of NH4Cl = 0.00476 M
With the new concentrations of NH3 and NH4Cl, we can calculate the new pH using the Henderson-Hasselbalch equation.
pH = pKa + log([A-]/[HA])
pH = 9.24 + log(0.00905/0.00476)
pH ≈ 9.24 + log(1.901)
Using a calculator, we find that log(1.901) ≈ 0.2785.
pH ≈ 9.24 + 0.2785
pH ≈ 9.52
So, the change in pH when 5.00 mL of 0.100 M HCl is added to 100.0 mL of the buffer solution is approximately -0.72. The final pH is 9.52.