Answer to Question #345050 in Chemistry for soph

Question

4. A solution of HF contains 12 grams in 500.0mL. (5 marks)

a. Find [H+]. Make sure to show the ionization equation (with water)

. b. Calculate the pH of the above solution.

5. Predict the products of the following neutralization reactions (be sure to balance the equation and include states as well). (2 marks) a. HCl(aq) + Mg(OH)2(s)  b. H3PO4(aq) + KOH(s) 

Answer 1

Solution (4):

The molar mass of HF is 20 g/mol

Therefore,

Moles of HF = (12 g HF) × (1 mol HF / 20 g HF) = 0.6 mol HF

Liters of solution = (500 mL) × (1 L / 1000 mL) = 0.5 L

Molarity = Moles of solute / Liters of solution

Therefore,

Molarity of HF solution = (0.6 mol) / (0.5 L) = 1.2 mol/L = 1.2 M

HF is a weak acid so it will not dissociate into its ion completely

K_a for HF is 7.2×10⁻⁴

The balanced equation for the ionization of HF is:

HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F⁻(aq)

or:

HF(aq) ⇌ H⁺(aq) + F⁻(aq)

Summarize the initial conditions, the changes, and the equilibrium conditions in the following ICE table:

According to the ICE Table:

[H⁺] = [F⁻] = x

[HF] = 1.2 − x

Substitute the equilibrium concentrations into the expression for the acid ionization constant K_a:

Solving for x we get:

x = 0.029

[H⁺] = 0.029 M

Calculate the pH as the negative of the base-10 logarithm of [H⁺]:

pH = −log[H⁺] = −log(0.029) = 1.5376 = 1.54

pH = 1.54

Answer (4): [H⁺] = 0.029 M; pH = 1.54

Solution (5):

(a): 2HCl(aq) + Mg(OH)₂(s) → MgCl₂(aq) + 2H₂O(l)

(b): H₃PO₄(aq) + 3KOH(aq) → K₃PO₄(aq) + 3H₂O(l)