Homework Help

If I have HCN+H2O<-->H+ + CN and Molarity is  [.5] and Ka of 4.93x10 ^-10  ...

user profile pic

mpumpkin | Salutatorian

Posted May 20, 2013 at 12:23 AM via web

dislike 1 like

If I have HCN+H2O<-->H+ + CN and Molarity is  [.5] and Ka of 4.93x10 ^-10

 

what would [H+] m [OH-], pH, pOH and % of ionization be.

 

 

 

 

1 Answer | Add Yours

user profile pic

jerichorayel | College Teacher | (Level 1) Senior Educator

Posted May 20, 2013 at 2:00 AM (Answer #1)

dislike 1 like

Since the water is part of the reactants, it would be better if we write it also in the product side as hydronium ion to avoid confusion. We can rewrite the equation to be:

`HCN + H_2O <=> H_3O^(+) + CN^(-)`

   0.5                          0                 0

   -x                            x                 x

-----------------------------------------------

    0.5 -x                      x                 x

`x =[H_3O^(+)] = [CN^(-)]`

`k_(a) = ([H_3O^(+)][CN^(-)])/([HCN])`

`k_(a) = ([x][x])/([0.5-x])`

`4.93x10^(-10) =([x^(2)])/([0.5-x])`

we assume that x <<<<< 0.5 therefore,

`4.93x10^(-10) =([x^(2)])/([0.5])`

`4.93x10^(-10) * (0.5)=(x^(2))`

`2.465e-10 = x^(2)`

`x = 1.57x10^(-5) =[H_3O^(+)] = [CN^(-)]`

`pH = -log (1.57x10^(-5)) =4.804`

`pOH = 14 - pH = 14 -4.804 =9.196`

`OH^(-) = 10^(-9.196) =6.368x10^(-10)`

 

`% ionization = x/([HCN]) * 100`

`% ionization = (1.57x10^(-5))/(0.5)`

`% ionization = 0.00314%`

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes