Calculate the change in pH when 4.00 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).
Delta pH=?
Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Delta pH=?
To calculate the change in pH when a solution is added to a buffer, we need to consider the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given as:
pH = pKa + log([A-]/[HA])
Where pH is the current pH of the buffer solution, pKa is the acid dissociation constant of the weak acid in the buffer (in this case, NH4+), [A-] is the concentration of the conjugate base (NH3), and [HA] is the concentration of the weak acid (NH4+).
Now let's calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of the buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).
1. Calculate moles of HCl added:
Moles of HCl = volume of HCl (in L) x concentration of HCl (in mol/L)
Moles of HCl = 4.00 mL x (1 L / 1000 mL) x 0.100 mol/L
Moles of HCl = 0.004 mol HCl
2. Calculate moles of NH4+ neutralized:
Moles of NH4+ = moles of HCl added (since HCl reacts with NH4+ to form NH3)
Moles of NH4+ = 0.004 mol NH4+
3. Calculate new concentrations of NH4+ and NH3 in the buffer solution:
Volume of buffer solution = volume of buffer (100.0 mL) + volume of HCl added (4.00 mL)
Volume of buffer solution = 104.0 mL = 0.104 L
New concentration of NH4+ = moles of NH4+ / volume of buffer solution
New concentration of NH4+ = 0.004 mol / 0.104 L
New concentration of NH4+ = 0.0385 M
New concentration of NH3 = initial concentration of NH3 (0.100 M)
4. Calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = pKa + log([NH3] / [NH4+])
pH = pKa + log(0.100 M / 0.0385 M)
The pKa value for NH4+ is 9.25.
Now, you can calculate the new pH by substituting the values into the equation and solving for pH.
Repeat the same steps to calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution. However, in this case, you will need to consider that NaOH is a strong base and will fully react with the NH4+ to form NH3. The reaction will create water and Na+ ions, changing the concentrations of NH4+ and NH3 in the buffer solution.
When you post these problems it would help if you gave the pKa or pKb since th number I look up in the book may not be the same as that in your text. I will use 1.75E-6 for Kb or NH3 which makes pKa = 9.24
millimoles NH3 = 100 x 0.1 = 10.
mmols NH4Cl = 100 x 0.1 = 10.
Initial pH = 9.24 + log(10/10) = 9.24.
When 4.00 mL of 0.1M HCl is added that adds 4.00 x 0.100 = 0.4 millimol acid.
........NH3 + H^+ ==> NH4^+
I.......10.0...0......10.0..
add acid.......0.400.......
C.....-0.40..-0.40...0.40...
E.......9.60...0.......10.4.....
Substitute the equil line into the HH equation and solve for pH.
Then subtract from the original pH to find the difference.
The addition of NaOH is done the same way. Don't forget that the difference will be + for one of them and - for the other one.