Answer to Question #344279 in Chemistry for soph

Question

11. Calculate [OH- ] ions in a 0.125 M solution of nitrous acid. Is this solution acidic or basic; how do you know? Make sure you include the ionization equation. (6 marks)

12. Calculate the [H3O+] in a 2.00 L solution of NaOH, a strong base, if it contains 0.800 g of solute. Is this solution acidic or basic; how do you know? Make sure you include the dissociation equation. (6 marks)

Answer 1

11. Calculate [OH⁻] ions in a 0.125 M solution of nitrous acid. Is this solution acidic or basic; how do you know? Make sure you include the ionization equation.

Solution (11):

Nitrous acid (HNO₂) is a weak acid (K_a = 4.5×10⁻⁴ for HNO₂).

The balanced equation for the ionization of HNO₂ is:

HNO₂(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NO₂⁻(aq)

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

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

Solving for x we get:

x = 0.00728

[H₃O⁺] = x = 0.00728 M

Calculate the pH as the negative of the base-10 logarithm of [H₃O⁺]:

pH = −log[H₃O⁺] = −log(0.00728) = 2.14

pH = 2.14, it is less than 7 so this solution is acidic

For any aqueous solution at 25^∘C:

[H₃O⁺][OH⁻] = K_w = 1.0×10⁻¹⁴

Therefore,

[OH⁻] = K_w / [H₃O⁺] = (1.0×10⁻¹⁴) / (0.00728) = 1.37×10⁻¹²

[OH⁻] = 1.37×10⁻¹² M

Answer (11): [OH⁻] = 1.37×10⁻¹² M; this solution is acidic

12. Calculate the [H₃O⁺] in a 2.00 L solution of NaOH, a strong base, if it contains 0.800 g of solute. Is this solution acidic or basic; how do you know? Make sure you include the dissociation equation.

Solution (12):

The molar mass of sodium hydroxide (NaOH) is 40 g/mol

Therefore,

Moles of NaOH = (0.800 g NaOH) × (1 mol NaOH / 40 g NaOH) = 0.020 mol NaOH

Molarity = Moles of solute / Liters of solution

Therefore,

Molarity of NaOH solution = (0.020 mol) / (2.00 L) = 0.01 mol/L = 0.01 M

Sodium hydroxide (NaOH) is a strong base because it almost completely ionized in water.

The balanced equation for the ionization of NaOH is:

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

And thus a 0.01 M solution of NaOH is stoichiometric in Na⁺ and OH⁻, i.e. [Na⁺] = [OH⁻] = 0.01 M

[OH⁻] = 0.01 M

For any aqueous solution at 25^∘C:

[H₃O⁺][OH⁻] = K_w = 1.0×10⁻¹⁴

Therefore,

[H₃O⁺] = K_w / [OH⁻] = (1.0×10⁻¹⁴) / (0.01) = 1.0×10⁻¹²

[H₃O⁺] = 1.0×10⁻¹² M

Calculate the pH as the negative of the base-10 logarithm of [H₃O⁺]:

pH = −log[H₃O⁺] = −log(1.0×10⁻¹²) = 12

pH = 12, it is more than 7 so this solution is basic

Answer (12): [H₃O⁺] = 1.0×10⁻¹² M; this solution is basic