There is a solution that is 3.5×10?2 M in HCN (Ka=4.9×10?10) and 1.6×10?2 M in N

Question

There is a solution that is 3.5×10?2 M in HCN (Ka=4.9×10?10) and 1.6×10?2 M in NaCN. Part A Calculate the concentrations of all species present in this solution. Express your answers in the given order using two significant figures separated by commas.

There is a solution that is 3.5×10-2 M in HCN (Ka 4.9 × 10-10) and 1.6×10-2 M in NaCN. Part A Calculate the concentrations of all species present in this solution Express your answers in the given order using two significant figures separated by commas. HO L CN L.HON OH Na]- [H3O+ ], [CN-], [HCN], [OH-], [Na+] = Submit My Answers Give Up

Explanation / Answer

NaCN -----------> Na+ +    CN-

0.016M                0.016                 0.016M

      CN- + H2O ---------> HCN + OH-

0.016                              0         0

-x                                  +x         +x

0.016-x                         +x         +x

Kb = [HCN][OH-]/[CN-]

Kb = Kw/Ka

     = 1*10-14/4.9*10-10 = 2.04*10-5

2.04*10-5 = x*x/0.016-x

32.64*10-8 = x2

x =5.71*10-4 M

[HCN] = x = 5.71*10-4 M

[OH-] = x = 5.71*10-4 M

Kw = [H3O+][OH-]

[H3O+] = Kw/[OH-]

            = 1*10-14/5.71*10-4 = 1.75*10-11 M

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