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5 April, 03:15

What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What percentage of the acid is dissociated?

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  1. 5 April, 03:36
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    a) pH = 4.213

    b) % dis = 2 %

    Explanation:

    Ch3COONa → CH3COO - + Na+

    CH3COOH ↔ CH3COO - + H3O+

    ∴ Ka = 1.8 E-5 = ([ CH3COO - ] * [ H3O + ]) / [ CH3COOH ]

    mass balance:

    ⇒ C CH3COOH + C CH3COONa = [ CH3COOH ] + [ CH3COO - ]

    ∴ C CH3COOH = 3.40 mM = 3.4 mmol/mL * (mol/1000mmol) * (1000mL/L)

    ∴ C CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

    ⇒ [ CH3COOH ] = 4.4 - [ CH3COO - ]

    charge balance:

    ⇒ [ H3O + ] + [ Na + ] = [ CH3COO - ] + [ OH - ] ... is negligible [ OH-], comes from water

    ⇒ [ CH3COO - ] = [ H3O + ] + 1.00

    ⇒ Ka = (([ H3O + ] + 1) * [ H3O + ]) / (3.4 - [ H3O+])) = 1.8 E-5

    ⇒ [ H3O + ]² + [ H3O + ] = 6.12 E-5 - 1.8 E-5 [ H3O + ]

    ⇒ [ H3O + ]² + [ H3O + ] - 6.12 E-5 = 0

    ⇒ [ H3O + ] = 6.12 E-5 M

    ⇒ pH = - Log [ H3O + ] = 4.213

    b) (% dis) * mol acid = C CH3COOH = 3.4

    ∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

    ⇒ % dis = 3.4 / 1.7 = 2 %
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