When 0.005 mol NaOH is added to 0.05 L of buffer solution (0.24 M NH3 and
0.20 M NH4Cl), what is the pH of the resulting solution?
Ka (NH4
+) = 5.6 x 10–1
To determine the pH of the resulting solution, we need to consider the reaction that occurs when NaOH is added to the buffer solution.
NaOH + NH4Cl -> NH3 + H2O + NaCl
Since NaOH is a strong base, it reacts completely with NH4Cl to form NH3 and NaCl. The NH3 reacts with water to form NH4+ and OH-. The NaCl remains dissociated in the solution.
So, after the reaction, we have NH4+ and OH- ions in the solution in addition to the NH3 from the original buffer solution.
The initial concentration of NH4Cl in the buffer solution is 0.20 M, and the volume of the buffer solution is 0.05 L. So the number of moles of NH4Cl is:
moles of NH4Cl = 0.20 M * 0.05 L = 0.01 mol
When 0.005 mol of NaOH is added, it reacts with an equal number of moles of NH4Cl to form NH3 and NaCl. So the remaining moles of NH4Cl in the solution is:
remaining moles of NH4Cl = 0.01 mol - 0.005 mol = 0.005 mol
Since NH4Cl is a weak acid, it partially dissociates into NH4+ and Cl- ions. The concentration of NH4+ ions is equal to the concentration of NH4Cl, which is 0.20 M.
Now, we need to calculate the concentration of OH- ions in the solution.
The reaction of NH3 with water can be represented as:
NH3 + H2O -> NH4+ + OH-
The Ka value for NH4+ (from the given information) is 5.6 x 10^-1.
Ka = [NH4+][OH-] / [NH3]
Since the concentration of NH4+ is 0.20 M (from NH4Cl), and the concentration of NH3 is 0.24 M (from the original buffer solution), we have:
5.6 x 10^-1 = (0.20 M)(OH-) / 0.24 M
Solving for [OH-], we get:
(0.20 M)(OH-) = (5.6 x 10^-1)(0.24 M)
(OH-) = (5.6 x 10^-1)(0.24 M) / 0.20 M
(OH-) = 0.672 x 10^-1 M
Now, we can calculate the concentration of OH- in moles per liter of solution:
[OH-] = (0.672 x 10^-1 M) * (0.05 L) = 0.0336 x 10^-1 mol/L
Since the concentration of OH- is equal to the concentration of H+ in a solution, the pOH of the solution is:
pOH = -log[OH-] = -log(0.0336 x 10^-1) = 1.47
Finally, we can calculate the pH of the resulting solution:
pH = 14 - pOH = 14 - 1.47 = 12.53
Therefore, the pH of the resulting solution is 12.53.