Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

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