Calculate the pH of:
a) ammonia, 0.1M, in water
b) ammonium chloride, 0.1M, in water
c) a buffer solution containing ammonia (0.1M)and ammonium chloride (0.01M)
For NH4^+, pKa = 9.25
a)
NH3 + H2O ==> NH4 + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
(NH4^+) = y
(OH^-) = y
(NH3) = 0.1 - y
Plug into Kb and solve for y.
b) NH4Cl + HOH ==> NH4^+ + Cl^-
The NH4^+ hydrolyzes as follows:
NH4^+ + H2O ==> NH3 + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+)
(NH3) = y
(H3O^+) = y
(NH3) = 0.1 - y
solve for y.
c) Use the Henderson-Hasselbalch equation.
Post your work if you get stuck on any of these.
To calculate the pH of a solution, you need to consider the dissociation of the given compound and the equilibrium between the acidic and basic forms. In this case, we will consider the dissociation of ammonia (NH3) into ammonium ions (NH4+) and hydroxide ions (OH-).
a) For ammonia (NH3) in water:
The dissociation of ammonia is given by the equation NH3 + H2O ⇌ NH4+ + OH-. Ammonia is a weak base, so it does not completely dissociate in water. To calculate the pH, we need to determine the concentration of OH- ions in the solution.
The concentration of OH- ions can be calculated using the equation Kw = [H+] [OH-], where Kw is the ionization constant for water (10^-14 M^2). Since ammonia is a weak base with a low concentration (0.1M), we can assume that the contribution of OH- from the autoionization of water is negligible.
Therefore, we can set up an ICE (Initial, Change, Equilibrium) table to determine the concentration of OH- ions:
NH3 + H2O ⇌ NH4+ + OH-
Initial: 0.1 M 0 M 0 M
Change: -x +x +x
Equilibrium: 0.1 - x x x
Since x is small compared to 0.1, we can approximate 0.1 - x as 0.1. Thus, [NH3] ≈ 0.1 M, [NH4+] ≈ x, and [OH-] ≈ x.
Now, we can use the relationship Kw = [H+] [OH-]. In this case, we assume that the concentration of [OH-] is approximately equal to the concentration of [NH4+], so we substitute [OH-] with x:
Kw = [H+] [NH4+]
10^-14 = [H+] * x
Since we know that pOH = -log[OH-] and pH + pOH = 14, we can rearrange the equation to solve for pH:
[H+] = 10^(-14) / x
pH = -log([H+]) = -log(10^(-14) / x) = -log(10^(-14) * (1/x)) = -(-14 * log(10) + log(1/x))
pH ≈ 14 + log(x)
Since x represents the concentration of NH4+, we can substitute x with 0.1 (assuming it fully dissociates):
pH ≈ 14 + log(0.1)
pH ≈ 14 + (-1) = 13
Therefore, the pH of a 0.1 M ammonia solution in water is approximately 13.
b) For ammonium chloride (NH4Cl) in water:
Ammonium chloride is the salt of a strong acid (HCl) and a weak base (NH4+). Salts of weak bases and strong acids are acidic in nature. To calculate the pH, we need to consider the dissociation of NH4Cl in water and the hydrolysis of NH4+.
NH4Cl dissociates into NH4+ and Cl-. However, NH4+ being a weak acid, undergoes hydrolysis as follows: NH4+ + H2O ⇌ NH3 + H3O+.
Since NH4+ is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([A-]/[HA])
pH = 9.25 + log ([NH3]/[NH4+])
Given the concentration of NH4Cl as 0.1 M, we can assume that [NH4+] = 0.1 M.
Therefore,
pH = 9.25 + log (0.1/0.1)
pH = 9.25 + log (1)
pH = 9.25 + 0
pH = 9.25
Thus, the pH of a 0.1 M ammonium chloride solution in water is approximately 9.25.
c) For the buffer solution containing ammonia (0.1 M) and ammonium chloride (0.01 M):
A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in equilibrium. In this case, we have ammonia (NH3) as the weak base and ammonium chloride (NH4Cl) as its conjugate acid.
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Given that pKa = 9.25, [A-] = concentration of NH3 = 0.1 M, and [HA] = concentration of NH4+ = 0.01 M, we can substitute these values into the equation:
pH = 9.25 + log (0.1/0.01)
pH = 9.25 + log (10)
pH = 9.25 + 1
pH = 10.25
Therefore, the pH of a buffer solution containing ammonia (0.1 M) and ammonium chloride (0.01 M) is approximately 10.25.