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

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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.

 

 

 

 

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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:

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