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:
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
