Determine the pH of each solution:

(a) 0.10 M NH4Cl

can someone show me how to do this one so i have the idea of how to do the rest. thanks!

To determine the pH of a solution, you need to understand the concept of an acid or a base and consider the dissociation of the solute in water.

In this case, NH4Cl is a salt of a weak base (NH3) and a strong acid (HCl). When NH4Cl dissolves in water, it dissociates into NH4+ (ammonium ion) and Cl- (chloride ion).

Since NH4+ is the conjugate acid of the weak base NH3, it can react with water to produce H3O+ ions (hydronium ions) and NH3 through the following equation:

NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)

The H3O+ ions contribute to the acidity of the solution, determining its pH.

To calculate the pH of the solution, you need to know the concentration of NH4Cl. In this case, it is given as 0.10 M NH4Cl.

1. Write out the dissociation equation:

NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)

2. Since NH4Cl is a strong electrolyte, it completely dissociates. So, the concentrations of NH4+ and Cl- ions are the same as the concentration of NH4Cl, which is 0.10 M.

3. Since NH4+ can react with water and produce H3O+ ions, the concentration of H3O+ ions at equilibrium is also 0.10 M.

4. Finally, use the formula for pH:

pH = -log[H3O+]

Substitute the concentration of H3O+ (0.10 M) into the formula:

pH = -log(0.10)

Calculating:

pH ≈ 1.00

Therefore, the pH of a 0.10 M NH4Cl solution is approximately 1.00.

The pH of salts is determined by the hydrolysis of the salt. In the case of NH4Cl, the NH4^+ is hydrolyzed because it is a stronger acid than H2O (the NH4^+ donates a H^+ to H2O).

.........NH4^+ + H2O ==> H3O^+ + NH3
.........0.1...............0......0
change....-x...............x......x
equil....0.1-x.............x......x

Ka for NH4^+ = (Kw/Kb for NH3) = (H3O^+)(NH3)/(NH4+)
Substitute from the ICE chart and solve for x which is H3O^+, and convert that to pH.