What is the pH of a 0.250 M NH4Cl solution? Kb for NH3 = 1.8 × 10−5.
Thank you.
Why did the NH4Cl solution go to therapy? Because it couldn't find its pH balance!
Okay, now let's get serious. NH4Cl is an acidic salt because NH4+ is the conjugate acid of a weak base, NH3. This means that NH4Cl will dissociate in water to produce NH4+ and Cl- ions. The NH4+ ion can act as an acid by donating a proton (H+) to water.
To find the pH of the solution, we need to determine the concentration of H+ ions. Since NH4Cl is a strong electrolyte, it will dissociate completely in water:
NH4Cl → NH4+ + Cl-
The concentration of NH4Cl is given as 0.250 M, so the concentration of NH4+ will also be 0.250 M. Since NH4+ is an acidic ion, we can assume it completely dissociates into H+ ions:
NH4+ → H+ + NH3
Therefore, the concentration of H+ ions is 0.250 M.
Now, let's use the Kb value for NH3 to calculate the concentration of OH- ions:
Kb = [NH3][OH-] / [NH4+]
Since the concentration of NH3 is equal to the concentration of NH4+, we can let x be the concentration of OH- ions. Thus:
Kb = x^2 / 0.250
But Kb for NH3 = 1.8 × 10^-5, so:
1.8 × 10^-5 = x^2 / 0.250
Solving for x, we find x = 7.5 × 10^-4.
Now, since the concentration of OH- is 7.5 × 10^-4 M, we can use the equation Kw = [H+][OH-] to find the concentration of H+:
Kw = (0.250 - x)(x)
Kw = (0.250 - 7.5 × 10^-4)(7.5 × 10^-4)
Simplifying this equation gives us a concentration of H+ ions of approximately 1.82 × 10^-11 M.
Finally, to find the pH, we can use the equation pH = -log[H+]:
pH = -log(1.82 × 10^-11)
After consulting with my comedy advisors, we've decided to skip the pH joke this time. The final pH of the NH4Cl solution is approximately 10.74.
Hope this helps, and keep those chemistry questions coming!
To find the pH of a solution that contains NH4Cl, we need to consider the hydrolysis reaction of NH4Cl in water. NH4Cl consists of the ammonium ion (NH4+) and the chloride ion (Cl-). While the chloride ion does not undergo hydrolysis, the ammonium ion does.
When the ammonium ion undergoes hydrolysis, it reacts with water to form ammonia (NH3) and hydronium ion (H3O+):
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant for this reaction is the base equilibrium constant, Kb. In this case, Kb = 1.8 × 10−5.
To find the pH, we need to find the concentration of hydronium ion (H3O+). Since the reaction is in equilibrium, the concentration of NH3 will be equal to the concentration of hydronium ion (assuming the system is at equilibrium).
Let's define x as the concentration of NH3 (and H3O+). Since NH4Cl dissolves completely and dissociates to give one NH4+ ion for every chloride ion (Cl-), the initial concentration of NH4+ is equal to the concentration of NH4Cl, which is 0.250 M.
Using the equilibrium constant expression, we have:
Kb = [NH3][H3O+] / [NH4+]
Plugging in the values:
1.8 × 10−5 = x^2 / 0.250
Rearranging the equation:
x^2 = 1.8 × 10−5 × 0.250
Solving for x:
x ≈ √(1.8 × 10−5 × 0.250)
x ≈ 0.00237 M
Since the concentration of NH3 (and H3O+) is equal to x, the concentration of hydronium ion (H3O+) is approximately 0.00237 M.
To find the pH, we can use the formula:
pH = -log[H3O+]
Plugging in the value:
pH ≈ -log(0.00237)
pH ≈ 2.62
Therefore, the pH of the 0.250 M NH4Cl solution is approximately 2.62.
To determine the pH of a solution of NH4Cl, we need to consider the ionization of NH4Cl in water.
NH4Cl dissociates into NH4+ and Cl- ions when dissolved in water. The NH4+ ion can also react with water to produce NH3 and H3O+ ions. This reaction is an acid-base reaction, with NH4+ acting as an acid and H2O acting as a base.
The Kb value for NH3 (ammonia) is given as 1.8 × 10−5. Kb is the equilibrium constant for the reaction of NH3 with water to form NH4+ and OH- ions:
NH3 + H2O ⇌ NH4+ + OH-
Since NH4+ is an acid, we can use the equation for the acid dissociation constant (Ka) to express the equilibrium constant in terms of the concentrations of NH4+ and H3O+ ions:
Ka = [NH4+][OH-] / [NH3]
Since Kb is the conjugate base constant for the reaction, we can use the relationship Ka × Kb = Kw (the ion product of water, equal to 1.0 × 10^-14 at 25°C).
Therefore, we have:
Ka × Kb = Kw
[NH4+][OH-] / [NH3] × Kb = Kw
[NH4+][OH-] / [NH3] = Kw / Kb
[NH4+][OH-] / [NH3] = (1.0 × 10^-14) / (1.8 × 10^-5)
Now, we need to consider the expression of Kw in terms of H3O+ and OH- concentrations:
Kw = [H3O+][OH-]
Substituting this into the equation, we have:
[NH4+] / [NH3] × [OH-] = (1.0 × 10^-14) / (1.8 × 10^-5)
Since the concentration of OH- ions is equal to the concentration of H3O+ ions in water (at equilibrium), we can simplify the equation further:
[NH4+] / [NH3] × [OH-] = (1.0 × 10^-14) / (1.8 × 10^-5)
Assuming NH4+ completely dissociates into NH3 and H3O+ ions, the concentration of NH4+ is equal to the initial concentration of NH4Cl in the solution:
[NH4+] = 0.250 M
Thus, we have:
(0.250 M) / [NH3] × [OH-] = (1.0 × 10^-14) / (1.8 × 10^-5)
Now, from the equation for NH4+ reacting with water (NH4+ + H2O ⇌ NH3 + H3O+), we can conclude that [NH3] = [OH-]. Therefore, we have:
(0.250 M) / [OH-]^2 = (1.0 × 10^-14) / (1.8 × 10^-5)
Rearranging the equation to solve for [OH-], we have:
[OH-]^2 = (0.250 M) × (1.8 × 10^-5) / (1.0 × 10^-14)
[OH-]^2 = 4.5 × 10^-6
Taking the square root of both sides, we find:
[OH-] = √(4.5 × 10^-6) = 6.71 × 10^-3 M
Since [OH-] = [NH3], the concentration of NH3 is also 6.71 × 10^-3 M.
Finally, using the equation for the pH of a solution, we have:
pOH = -log [OH-] = -log (6.71 × 10^-3) ≈ 2.17
And since pH + pOH = 14, we can find the pH:
pH = 14 - pOH = 14 - 2.17 ≈ 11.83
Therefore, the pH of a 0.250 M NH4Cl solution is approximately 11.83.