Determine the pH of 0.18M NH4Cl.
NH4^+ + H2O ==> H3O^+ + NH3
Ka = (Kw/Kb) = (NH3)(H3O^+)/(NH4^^+)
Set up an ICE chart and solve for (H3O^+), then convert to pH.
To determine the pH of a solution of NH4Cl, we need to know the dissociation of the compound in water. NH4Cl dissociates into NH4+ and Cl- ions.
NH4+ is the conjugate acid of NH3 (ammonia), which can act as a weak base. It can accept a proton (H+) to form NH3.
Since NH4+ is a weak acid, it undergoes a partial dissociation in water. This means that only a fraction of NH4+ ions will react with water to produce NH3 and H+ ions.
The Cl- ion is the conjugate base of HCl (hydrochloric acid), which is a strong acid and completely dissociates in water to produce H+ ions.
Given that the concentration of the NH4Cl solution is 0.18 M, we can assume that the concentration of NH4+ and Cl- ions is initially 0.18 M.
Since the Cl- ion comes from a strong acid, we can ignore its contribution to the pH calculation.
Thus, we only need to consider the contribution of NH4+.
The dissociation reaction is as follows:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H+(aq)
The equilibrium constant expression, Ka, for this reaction can be written as:
Ka = [NH3][H+]/[NH4+]
Since we have assumed that the concentration of NH4+ is 0.18 M at the start, and we do not have any information about the equilibrium concentrations, we can consider all of NH4+ to have dissociated into NH3 and H+. This means that the concentration of NH4+ ions will be zero (0 M) at equilibrium.
Thus, the [NH4+] in the Ka expression becomes zero, and the Ka expression simplifies to:
Ka = [NH3][H+]
We can approximate [NH3] as the initial concentration of NH4+ since the dissociation of NH4+ is only partial. Therefore, we have:
Ka = [NH4+][H+]
We can substitute the known value for Ka of NH4+ as a weak acid, which is approximately equal to 5.8 x 10^-10.
5.8 x 10^-10 = (0.18)([H+])
To find [H+], we rearrange the equation:
[H+] = (5.8 x 10^-10)/(0.18)
Calculating this value, we get:
[H+] = 3.2 x 10^-9 M
The pH of a solution can be calculated using the equation:
pH = -log[H+]
So, plugging in the value of [H+], we have:
pH = -log(3.2 x 10^-9)
pH ≈ 8.5
Therefore, the pH of a 0.18 M NH4Cl solution is approximately 8.5.
To determine the pH of a solution of NH4Cl, you need to consider the dissociation of NH4Cl in water. NH4Cl is a salt that consists of a weak acid (NH4+) and a strong base (Cl-).
Step 1: Write the balanced equation for the dissociation of NH4Cl in water:
NH4Cl -> NH4+ + Cl-
Step 2: NH4+ can act as a weak acid by donating a proton (H+) to the water, according to the equation:
NH4+ + H2O -> NH3 + H3O+
Step 3: Since NH4+ is a weak acid, it will only partially dissociate in water. This means that not all of the NH4+ ions will donate a proton. Therefore, you need to calculate the concentration of the NH4+ ions that have dissociated.
Step 4: The concentration of NH4+ ions can be determined using the initial concentration of NH4Cl. In this case, the initial concentration is given as 0.18 M.
Step 5: Since NH4Cl is a 1:1 electrolyte, the concentration of NH4+ ions that have dissociated will also be 0.18 M.
Step 6: Now, determine the concentration of H3O+ ions formed from the dissociation of NH4+ by multiplying the concentration of the dissociated NH4+ ions by the dissociation constant of NH4+ (Ka for NH4+). The Ka value for NH4+ is 5.6 x 10^-10.
Step 7: [H3O+] = (0.18 M) x (5.6 x 10^-10)
Step 8: Calculate the pH using the formula pH = -log[H3O+].
By performing the calculations, you will find the pH of the 0.18 M NH4Cl solution.