How do you calculate the pH when the following substances are added to a buffer.(all of the solutions are at .10M conc.)
The buffer: 50mL NH3 + 50mL NH4NO3
1) 10mL buffer + 6mL water
2)10mL buffer + 5mL water +1mL HCl
3)10mL buffer + 6mL HCl
4)10mL buffer + 5mL water + 1mL NaOH
I'm really confused on how the equations are set up for these types of problems (buffer solns).
Thanks
fgs
To calculate the pH of a buffer solution, you need to consider the equilibrium reactions that occur between the acid and its conjugate base in the presence of water.
In this case, the buffer solution consists of NH3 (ammonia) and NH4NO3 (ammonium nitrate). When NH3 reacts with water, it forms NH4+ and OH- ions according to the equation:
NH3 + H2O ⇌ NH4+ + OH-
First, let's determine the initial concentrations of NH3, NH4+, and OH- in the buffer solution.
NH3:
Concentration of NH3 in 50 mL = 0.10 M * (50 mL / 100 mL) = 0.05 M
NH4+:
Concentration of NH4+ in 50 mL = 0.10 M * (50 mL / 100 mL) = 0.05 M
OH-:
Concentration of OH- in 50 mL of water = 10^-7 M (at room temperature)
Now, let's go through each scenario and calculate the resulting pH:
1) 10 mL buffer + 6 mL water:
Since water does not react with the components of the buffer, the concentrations of NH3, NH4+, and OH- will remain the same. Therefore, the pH will stay constant.
2) 10 mL buffer + 5 mL water + 1 mL HCl:
HCl is a strong acid that dissociates completely in water, so it will donate H+ ions to the solution. In this case, the concentration of H+ ions will increase.
H+:
Concentration of H+ from 1 mL of HCl = 0.10 M * (1 mL / 100 mL) = 0.001 M
Using the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid in the buffer solution.
NH3 acts as the base (A-) and NH4+ acts as the conjugate acid (HA). The pKa of NH4+/NH3 is around 9.25.
pH = 9.25 + log ([0.05 M]/[0.05 M + 0.001 M])
3) 10 mL buffer + 6 mL HCl:
Similar to scenario 2, the HCl will increase the concentration of H+ ions.
H+:
Concentration of H+ from 6 mL of HCl = 0.10 M * (6 mL / 100 mL) = 0.006 M
pH = 9.25 + log ([0.05 M]/[0.05 M + 0.006 M])
4) 10 mL buffer + 5 mL water + 1 mL NaOH:
NaOH is a strong base that dissociates completely in water, so it will react with H+ ions.
OH-:
Concentration of OH- from 1 mL of NaOH = 0.10 M * (1 mL / 100 mL) = 0.001 M
H+ + OH- ⇌ H2O
The OH- will react with the H+ ions present in the buffer solution, leading to a decrease in H+ concentration.
Using the Henderson-Hasselbalch equation again:
pH = 9.25 + log ([0.05 M]/[0.05 M + 0.001 M])
Note: The Henderson-Hasselbalch equation assumes that the concentration of water is constant and can be neglected, so the dilution of the solution does not significantly affect the pH.