Acid dissociation is an equilibrium Set up an ice table for

Acid dissociation is an equilibrium. Set up an ice table for the following reaction Ka HAen If the starting concentration of HA a 1.00 x 10-3 M solution has a pH-6.53, determine the Kfor the acid 2 of 2

Solution

Solution

pH = 6.53 = -log [H⁺]

[H⁺] = 10^-6.53 = 2.95 x 10^-7 M

At equilibrium [H₃O⁺] = 2.95 x 10^-7 M = ([A^-]

HA_{(aq)                         +}

H₂O_(l)                                =

H₃O⁺(aq)              +                                

A^-_(aq)

I

1.00 x 10^-3

0

0

C

-2.95 x 10^-7 M

+2.95 x 10^-7 M

+2.95 x 10^-7 M

E

(1.00 x 10^-3 -2.95 x 10^-7) M

2.95 x 10^-7 M

2.95 x 10^-7 M

pH = pKa + ln([A^-]/[HA])

6.53 = pKa + ln(2.95 x 10^-7 M/(1.00 x 10^-3 -2.95 x 10^-7)

6.53 = pKa + ln(2.95 x 10^-7/(1.0 x 10^-4)

6.53 = pKa + ln(0.00295)

6.53 = pKa -5.83

pKa = 6.53 + 5.83 = 12.36

pKa = -logKa

Ka = 10^-pKa = 10^-12.36

Ka = 4.36 x 10^-13

	HA(aq)                         +	H2O(l)                                =	H3O+(aq)              +	A-(aq)
I	1.00 x 10-3		0	0
C	-2.95 x 10-7 M		+2.95 x 10-7 M	+2.95 x 10-7 M
E	(1.00 x 10-3 -2.95 x 10-7) M		2.95 x 10-7 M	2.95 x 10-7 M