A 50.0 mL sample of buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate. If 5.55 mL of 0.092 M NaOH is added to this sol

Question

A 50.0 mL sample of buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate. If 5.55 mL of 0.092 M NaOH is added to this solution, identify the resulting number of moles of acetic acid, sodium acetate, and NaOH.

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Thiên Di 3 years 2021-08-31T04:44:04+00:00 1 Answers 114 views 0

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  1. thucuc
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    2021-08-31T04:45:27+00:00
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    Answer:

    see calculations below

    Explanation:

    Given:     HOAc     ⇄    [H⁺]     +    [OAc⁻]            

    C(i)          0.12M             0M           0.15M

    mix  => 5.55ml(0.092M NaOH) / (50ml + 5.55ml)

                = 0.00555(0.092)mole NaOH / 0.0555 L Soln

                = 0.0092M in NaOH is added into the initial buffer solution

                = 0.0092M in OH⁻ (NaOH is a strong base => 100% ionized)

    Rxn => Addition of 0.0092M OH⁻ will react with 0.0092M H⁺ shifting buffer            .           equilibrium to the right decreasing [HOAc] and increasing [OAc⁻] by    .           0.0092M each.

    Therfore …

    Given:           HOAc      ⇄    [H⁺]     +       [OAc⁻]            

    C(i)                0.12M             0M               0.15M

    ΔC           – 0.0092M           +x            +0.0092M

    C(f)             0.1108M             x                0.1592M  =>  New Concentrations            .                                                                                     after adding 0.0092M             .                                                                                     NaOH

    Substituting new acid and ion concentrations into Ka expression …

    Ka = [H⁺][OAc⁻]/[HOAc] = (x)(0.1592M)/(0.1108M) = 1.75 x 10⁻⁵M

    => x = [H⁺](new) = (1.75 x 10⁻⁵M*)(0.1108M)/(0.1592M) = 1.22 x 10⁻⁵M in H⁺ ions

    *units of Ka are Molar

    FYI => Adding a strong base to a buffer solution will shift pH to more basic.

              Adding a strong acid   to a buffer solution will shift pH to more acidic.

    => (such is a good way to check that your buffer calculations are correct.)

    NOTE => Question asks for moles of HOAc, Na⁺OAc⁻ & NaOH after adding base. Giving answers in terms of Molarity (moles/Liter) is same as moles. Therefore …

    [HOAc] = 0.1108M

    [NaOAc] = 0.1592M

    [NaOH] = ∅M (from rxn of H⁺ + OH⁻ => H₂O, all NaOH was consumed in acid/base reaction.  Remaining are only Na⁺ as a spectator ion and OH⁻ as a function of the new concentration of H⁺ => [OH⁻] = Kw/[H⁺] = 1 x 10⁻¹⁴/1.22 x 10⁻⁵ = 8.2 x 10⁻¹⁰M.

    Hope this helps. 🙂

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