What is the pH of 0.1 mol of NH4CL
I assume you meant 0.1 M NH4Cl or 0.1 mol/L NH4Cl
The NH4^+ is hydrolyzed this way
..................NH4^+ + H2O ==> NH3 + H3O^+
I...................0.1..........................0...........0
C......................-x........................x............x
E..................0.1-x........................x............x
Ka for NH4Cl = (Kw/Kb for NH3) = (x)(x)/(0.1-x)
Solve for x = (H3O^+) then convert to pH = -log(H3O^+)
Post your work if you get stuck.
To find the pH of a solution, we need to know the concentration of the hydronium ion (H3O+) in the solution. Since NH4Cl is a strong electrolyte that completely dissociates into NH4+ and Cl- ions in water, we can consider the reaction as NH4Cl → NH4+ + Cl-.
Since NH4+ is a weak acid, it can undergo a partial dissociation in water based on the equilibrium equation NH4+ + H2O ⇌ H3O+ + NH3. The concentration of NH4+ ion is equal to the initial concentration of NH4Cl, which is 0.1 mol.
Now, we need to determine whether NH4+ ion has a significant effect on the pH of the solution. To do this, we need the dissociation constant for NH4+ (Ka), which measures the strength of an acid.
The Ka for NH4+ is 5.6 x 10^-10.
Since NH4Cl is a salt consisting of NH4+ and Cl- ions, it undergoes hydrolysis in water: NH4+ + H2O ⇌ H3O+ + NH3.
Since NH4+ is a weak acid, it donates a proton, resulting in the formation of H3O+ ions. This causes an increase in the concentration of H3O+ ions in the solution, making it acidic.
To calculate the concentration of H3O+ ions and the pH, we need to consider the equilibrium expression for the hydrolysis reaction:
[ NH3 ][ H3O+ ]/ [ NH4+ ] = Ka
Given that the initial concentration of NH4+ is 0.1 mol and NH3 concentration is negligible compared to the NH4+ concentration, we can simplify the equation to:
[ H3O+ ] = [ NH4+ ] × sqrt(Ka)
Substituting the values:
[ H3O+ ] = 0.1 mol × sqrt(5.6 x 10^-10)
= 0.1 × 7.48 x 10^-6
= 7.48 x 10^-7 mol/L
Finally, to find the pH, we take the negative logarithm of the H3O+ concentration:
pH = -log ([ H3O+ ])
= -log (7.48 x 10^-7)
= 6.12
Therefore, the pH of the 0.1 mol NH4Cl solution is approximately 6.12.
To determine the pH of a solution, we need to consider the dissociation of NH4Cl in water.
NH4Cl dissociates into NH4+ and Cl- ions in water. NH4+ is a weak acid that can donate a proton (H+) while Cl- is a spectator ion and does not affect the pH of the solution.
The dissociation of NH4+ can be represented as follows:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
The equilibrium expression for this dissociation can be written as:
Ka = [NH3][H+]/[NH4+]
To calculate the pH, we need to consider the concentration of NH4+ and the equilibrium constant (Ka) for the dissociation of NH4Cl.
Given that we have 0.1 mol of NH4Cl, the concentration of NH4+ is also 0.1 mol.
The equilibrium constant (Ka) for the dissociation of NH4+ is approximately 5.6 x 10^-10 at 25°C.
Now we can use the equation of the Ka expression to determine the concentration of [H+].
Ka = [NH3][H+]/[NH4+]
5.6 x 10^-10 = [NH3][H+]/0.1
Since NH3 is a weak base, we can assume its concentration will be negligible compared to [NH4+], so we can cancel it out.
5.6 x 10^-10 ≈ [H+]/0.1
[H+] ≈ 5.6 x 10^-11 mol/L
To convert this concentration of H+ to pH, we can use the equation pH = -log[H+].
pH = -log(5.6 x 10^-11) ≈ 10.75
Therefore, the pH of a 0.1 mol NH4Cl solution is approximately 10.75.