Question

Consider the following dissociation of the weak acid, HCN: HCN(aq) ⇌ CN−(aq) + H+(aq) K = 6.2×10-10 A solution is made with an initial concentration of 2.60 M HCN. At equilibrium, what is the concentration of H+ ions in solution? (Hint: You may use the 5% approximation.)

?M

Answer 1

consider the dissociation of HCN

HCN ---> CN- + H+

K = [CN-] [H+] / [HCN]

now

using ICE table

initial conc of HCN , CN- , H+ are 2.6 , 0 , 0

change in conc of HCN , CN- , H+ are -x , x , x

so

equilibrium conc of HCN ,CN- , H+ are 2.6 - x , x , x

now

K = [CN-] [H+] / [HCN]

so

K = [x] [x] / [2.6 -x]

6.2 * 10-10 = x2 / ( 2.6-x)

as K is very very small , dissociation will be very less

so

2.6 >>>>> x  

2.6 -x ---> 2.6

so

6.2 * 10-10 = x2 / 2.6

x2 = 6.2 * 10-10 * 2.6

x = 4 * 10-5

now

[H+]eq = x = 4 * 10-5

so

the conc of H+ ions in solution at equilibrium is 4 * 10-5