An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid, HC2H3O2(aq), and 50. mL of 0.20 M sodium acetate, Na

Question

An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid,
HC2H3O2(aq), and 50. mL of 0.20 M sodium acetate, NaC2H3O2(aq). A 5.0 mL sample of 0.10 M NaOH(aq) is added to the buffer solution.
Which of the following is a correct pairing of the acetate species present in greater concentration and of the pH of the solution after the NaOH(aq) is added? (The pKa of acetic acid is 4.7.)
Acetate Species — pH
(A) HC2H3O2 — < 4.7
(B) HC2H3O2 — > 4.7
(C) C2H3O2 — < 4.7
(D) C2H3O2 — > 4.7

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Linh Đan 3 years 2021-08-31T22:34:00+00:00 1 Answers 86 views 0

Answers ( )

  1. kimchi2
    0
    2021-08-31T22:35:26+00:00
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    Answer:

    The answer is “Option B“.

    Explanation:

    \to CH_3COOH + NaOH \longleftrightarrow CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\ NaOH = (5 \ E-3\ L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\

    \to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\ L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\

    \to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\

    \to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7

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