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r a weak acid with a dissociation co assumption that the chai ssocatconstant of

ID: 303671 • Letter: R

Question




r a weak acid with a dissociation co assumption that the chai ssocatconstant of 1.00 x 10. Evaluate the equnbrium calculation by cf the weak acid can be ignored in an equilibrium with and without the calculating the hydronium ion concentration [HO'1 at [HA] listed below. assumption for initial concentrations of the weak acid (For the percent error calculation uadratic equation) as the correet' answer) value obtained without the assumption (from the HA Ratio of HA/Ka % error INa-- 1-assumption assumption 0.10 M 1.0 x 102 M 1.0x 10M 1.0 x 104 NM 1.0 x 10 M a reasonable value for the ratio of acid concentration to After filling out the table, identify equilibrium constant for the assumption to be valid.

Explanation / Answer

let the weak acid be represented as HA. It undergoes ionization in presence of water as

HA+ H2O ------------>A- + H3O+

Ka= [A-][H3O+]/[HA]= 1*10-5,

preparing the ICE table,

component                  initial concentration (M)            Change                  Eq. concentration

HA                                         M                                         -x                                M-x

A-                                           0                                          x                                  x

H3O+                                     0 x x

Ka= x2/(M-x)= 1*10-5   (1)

x2 = 1*10-5*(M-x) , if x cannot be neglected

x2= 1*10-5M- 1*10-5x

x2+1*10-5x -1*10-5M = 0 (2)

this is a quadratic equaiton of the form ax2+bx+c=0 whose roots as

x= {-b+-sqrt(b2-4ac)}/2a, here a= 1, b =-1*10-5 and c= -1*10-5

the +ve root need to be considered for this quadratic equation, hence x= {-b+sqrt(b2-4ac)}/2a

x= {-1*10-5 +sqrt(1*10-10+4*10-5M}/2a   (3)

when x is very very small, it is reasonable to assume that M-x is approximately equal to M.

hence Eq.1 becomes x2= 1*10-5M, x= sqrt(1*10-5M) (4)

% error = 100*(Eq.4-Eq.3)/Eq.3

x =[H3O+]

one sample calculation is shown and rest of them are tabulated. for

M= 1M, [HA]/Ka= 1/1*10-5 = 105, when [HA] =1 M

From Eq.3, x={ -1*10-5+sqrt(1*10-10+4*1*10-5M)}/2= 0.003157

from Eq.4, x= sqrt(1*10-5)= 0.00316278M

% error= 100*(0.00316278-0.003157)/0.003157=0.158

Rest of the calculations for different initial concentration of M is shown below

the data suggests that at lower concentration of M, the assumption of M-x approximatelty equal to M is not justified.