Question

Calculate the concentrations of all species in a 0.310-M solution of H2X if K1 = 3.16E-05 and K2 = 2.34E-08

[H2X] =

[H3O1+] =

[HX1-] =

[X2-] =

Answer 1

[H2X ] = 0.3069M

[ HX- ] = 0.0031M

[ X2- ] = 8.51×10^-6M

[ H3O+ ] = 0.0031M

Explanation

First dissociation

H2X + H2O <---------> HX- + H3O+

K1 = [ H3O+ ] [ HX- ]/ [ H2X] = 3.16 ×10^-5

at equillibrium

[ H2X ] = 0.310 - X

[ H3O+ ] = X

[ HX- ] = X

Therefore,

X^2/0.310 - X = 3.16 ×10^-5

we can assume 0.310 - X = 0.310

X^2 = 9.80 × 10^-6

X = 0.0031

[ H3O+ ] = 0.0031M

[HA-] = 0.0031M

[ H2X ] = 0.310 - 0.0031=0.3069M

second dissociation

HX- + H2O <-----------> X2- + H3O+   

K2 = [ X2- ] [ H3O+ ]/[ HX- ] = 2.34×10^-8

X^2 / (0.0031 - X ) = 2.34 ×10^-8

we can assume 0.0031 - x = 0.0031

X^2/0.0031 = 2.34 ×10^-8

X^2 = 7.25×10^-11

X = 8.51× 10^-6

Therefore,

[ X2- ] = 8.51× 10^-6M

[ H3O+ ] = 8.51 × 10^-6M

[HX- ] = 0.0031 - 0.00000851= 0.0031M

Total [ H3O+ ] = 0.0031M + 0.00000851= 0.0031M