calculate the pH of a 0.100M NH4Cl solution.

To calculate the pH of a solution of NH4Cl, we need to determine the concentration of the NH4+ and Cl- ions, as well as the equilibrium constant for the dissociation of water.

NH4Cl dissociates completely in water to form NH4+ and Cl- ions:

NH4Cl (s) --> NH4+ (aq) + Cl- (aq)

Since NH4Cl is a strong electrolyte, it dissociates completely, and we can consider NH4+ as a strong acid and Cl- as a spectator ion.

Now, let's construct the equation for the dissociation of water:

H2O (l) --> H+ (aq) + OH- (aq)

The equilibrium constant expression for the ionization of water is given by:

Kw = [H+][OH-] = 1.0 x 10^-14

Since the dissociation of NH4+ can affect the concentration of H+, we need to consider this as well. NH4+ acts as a weak acid and dissociates in water according to the following equation:

NH4+ (aq) + H2O (l) --> H3O+ (aq) + NH3 (aq)

The equilibrium constant expression for this reaction is given by:

Ka = [H3O+][NH3]/[NH4+]

Given that NH4Cl is a strong electrolyte, complete dissociation occurs, and the concentration of NH4+ is equal to the concentration of NH4Cl, which is 0.1 M.

Now, let's calculate the concentration of H3O+ ions:

[H+] = [H3O+] = [NH4+] = 0.1 M

Since [H+] = [H3O+], we can substitute [H3O+] with [H+] in the equation for Ka:

Ka = [H+][NH3]/[NH4+]

Substituting the known values, we get:

Ka = (0.1) * (0.1) / (0.1)

Ka = 0.1

Now, we can calculate the concentration of OH- ions using the Kw expression:

Kw = [H+][OH-] = 1.0 x 10^-14

Substituting [H+] = [NH4+] = 0.1 M, we get:

(0.1)([OH-]) = 1.0 x 10^-14

[OH-] = (1.0 x 10^-14) / 0.1

[OH-] = 1.0 x 10^-13 M

Now, we can use the concentration of H+ or OH- ions to find the pH or pOH of the solution, respectively.

Since pH + pOH = 14, we can calculate the pOH:

pOH = -log10([OH-])

pOH = -log10(1.0 x 10^-13)

pOH ≈ 12.0

Finally, we can calculate the pH:

pH = 14.0 - pOH

pH ≈ 14.0 - 12.0

pH ≈ 2.0

Therefore, the pH of a 0.100 M NH4Cl solution is approximately 2.0.

To calculate the pH of a solution, you need to consider the dissociation of the solute and the subsequent formation of ions in the solution.

NH4Cl is a salt that dissociates in water, forming NH4+ (ammonium) and Cl- (chloride) ions. The ammonium ion is a weak acid that can react with water to produce H3O+ ions. This reaction can be represented as follows:

NH4+ (aq) + H2O (l) -> NH3 (aq) + H3O+ (aq)

To calculate the pH of the solution, you need to determine the concentration of H3O+ ions produced by the reaction. This can be done using the equilibrium constant expression (Ka) for the dissociation of the ammonium ion. The Ka expression for the reaction is:

Ka = [NH3] [H3O+] / [NH4+]

Assuming that the dissociation of NH4+ is complete and [NH4+] can be considered the initial concentration of NH4Cl (0.100 M), and that the concentration of NH3 formed is small compared to the initial concentration, we can simplify the equation to:

Ka = [H3O+]

Now we can calculate the H3O+ concentration by using the value of Ka for the ammonium ion. The Ka value for NH4+ is 5.6 x 10^-10 at 25°C. Taking the square root of Ka gives us the concentration of H3O+:

[H3O+] = √(5.6 x 10^-10) = 7.48 x 10^-6 M

Finally, we can calculate the pH of the solution using the formula:

pH = -log[H3O+]

pH = -log(7.48 x 10^-6)
pH ≈ 5.13

Therefore, the pH of a 0.100 M NH4Cl solution is approximately 5.13.