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22 May, 21:23

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.

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  1. 22 May, 22:21
    0
    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
  2. 22 May, 23:45
    0
    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
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