ning.com/ibisoms/mod/ibis/view.php?id 3202986 CHEM332-spring17-GooD I ActMtes an

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ning.com/ibisoms/mod/ibis/view.php?id 3202986 CHEM332-spring17-GooD I ActMtes and Due Dates I ch9 1711-SSPM A 33h00 Print Cacuator Periodic Table Question 7 of g Sapling Learning Use activity coefficients to find the concentration of hydrogen ions in a solution of 65.0 mM butanoic acid (CsHTCOOH) and 0.1 MNaCl. The ionization constant of butanoic acid is Ka 1.52 x108. Take the size of C3HtCOO to be 500 pm. Activity coefficients can be found in this table. What is the pH of this solution? Include activity coefficients in your calculation. pH What is the fraction of dissociation of butanoicacidin this solution? Number

Explanation / Answer

Calculate activity coefficient, H+ and pH of butanoic acid solution

ionic strength (I) = 1/2CiZi^2

                           = 1/2(0.1 x 1^2 + 0.1 x 1^2)

                           = 0.1 M

Using Debye-Huckel relation for activity coefficient Y,

r for CH3COO- = 500 pm = 0.5 nm

Y[CH3COO-] = inv.log[(-0.51 x Zi^2 x sq.rt.(I))/(1+3.3 x r x sq.rt.(I))]

                      = inv.log[(-0.51 x 1 x sq.rt.(0.1))/(1+3.3 x 0.5 x sq.rt.(0.1))]

                      = 0.783

r for H3O+ = 900 pm = 0.9 nm

Y[H3O+] = inv.log[(-0.51 x Zi^2 x sq.rt.(I))/(1+3.3 x r x sq.rt.(I))]

               = inv.log[(-0.51 x 1 x sq.rt.(0.1))/(1+3.3 x 0.9 x sq.rt.(0.1))]

               = 0.826

So,

Ka = 1.52 x 10^-8 = (x^2)(0.783 x 0.826)/0.065

x = [H+] = 3.91 x 10^-5 M

pH = -log[H+] = 4.41

fraction of dissciation = 3.91 x 10^-5 x 100/0.065 = 0.06%

Calculate pH of 0.0035 M acetic acid

CH3COOH + H2O <==> H3O+ + CH3COO-

let x amount of acid has dissociated

Ka = [H3O+][CH3COO-]/[CH3COOH]

1.78 x 10^-5 = x^2/(0.0035 - x)

x^2 + 1.78 x 10^-5x - 6.23 x 10^-8 = 0

x = [H3O+] = [H+] = 2.41 x 10^-4 M

pH = -log[H+] = 3.62

Calculate pH of monoprotic acid

Ka = 7.4 x 10^-6

initial concentration = 0.38 M

HA + H2O <==> H3O+ + A-

let x amount of acid has dissociated

Ka = [H3O+][A-]/[HA]

7.4 x 10^-6 = x^2/0.38

x = [H3O+] = 1.677 x 10^-3 M

pH = -log[H3O+] = 2.775

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