write out the acid dissociation reaction for hydrochloric acid. b) calculate the pH of a solution of 5.0 x 10^-4 M HCl. c) write out the acid dissociation reaction for sodium hydroxide. d) calculate the pH of a solution of 7.0 x 10^-5 M NaOH.

a) HCl (g) + H₂O (l) → H + (aq) + Cl - (aq)

b) pH = 3.3

c) NaOH (s) + H2O (l) → Na + (aq) + OH - (aq)

d) pH = 9.85

Explanation:

Step 1: write out the acid dissociation reaction for hydrochloric acid

When HCl molecules dissolve in water, they dissociate into H + ions and Cl - ions. HCl is a strong, this meansit dissociates (almost) completely.

HCl (g) + H₂O (l) → H + (aq) + Cl - (aq)

b) calculate the pH of a solution of 5.0 x 10^-4 M HCl.

pH HCl = - log [HCl]

pH HCl = - log [5.0 * 10^-4)

pH = 3.3

c) write out the acid dissociation reaction for sodium hydroxide.

A strong base like sodium hydroxide (NaOH) will also dissociate completely into water; if you put in 1 mole of NaOH into water, you will get 1 mole of hydroxide ions.

NaOH (s) + H2O (l) → Na + (aq) + OH - (aq)

d) calculate the pH of a solution of 7.0 * 10^-5 M NaOH.

[OH-] = 7.0 * 10^-5 M

pOH = - log[OH-]

pOH = - log[7.0 * 10^-5 M]

pOH = 4.15

pH = 14 - pOH

pH = 14 - 4.15 = 9.85

The answer to your question is below

Explanation:

a) Acid dissociation reaction for HCl

HCl (g) + H₂O (l) ⇒ H⁺¹ (aq) + Cl⁻¹ (aq)

b) pH = - log [H⁺¹]

Concentration = 5 x 10⁻⁴

pH = - log [5 x 10⁻⁴]

pH = 3.3

c) Acid dissociation reaction

NaOH (s) + H₂O ⇒ Na⁺¹ (aq) + OH⁻¹ (aq)

d) pH

pOH = - log [7 x 10⁻⁵]

pOH = 4.2

pH = 14 - 4.2

pH = 9.8