If  50ml  of  0.2MNaCN  is mixed with  50ml  of  0.2MHCl , then  (Kb  for  CN−  =  2 × 10−5)  calculate concentration of  [H3O+] in terms of molarity  ×10−6

The reaction that occurs when NaCN is mixed with HCl is:

NaCN + HCl -> NaCl + HCN

Since the molarities and volumes of NaCN and HCl are the same, they will react in a 1:1 ratio and all of the NaCN and HCl will react to form NaCl and HCN.

The concentration of HCN formed is 0.2M (since all of the NaCN and HCl react to form HCN).

HCN is a weak acid and will partially ionize in water to form H3O+ and CN-. The ionization of HCN can be represented by the equation:

HCN + H2O -> H3O+ + CN-

The Kb for CN- is given as 2 x 10^-5. The Kb is related to the Ka (acid dissociation constant) of the conjugate acid (HCN) by the equation:

Kw = Ka x Kb

Where Kw is the ion product of water, which is 1 x 10^-14 at 25 degrees Celsius.

We can solve for Ka:

Ka = Kw / Kb = (1 x 10^-14) / (2 x 10^-5) = 5 x 10^-10

The ionization of HCN can be represented by the following ICE (Initial, Change, Equilibrium) table:

Initial: 0.2M, 0, 0 Change: -x, +x, +x Equilibrium: 0.2-x, x, x

Where x is the concentration of H3O+ formed.

We can set up the Ka expression:

Ka = [H3O+][CN-] / [HCN] = (x)(x) / (0.2 - x) = 5 x 10^-10

Assuming x is small compared to 0.2, the equation simplifies to:

x^2 / 0.2 = 5 x 10^-10

Solving for x gives:

x = sqrt((0.2)(5 x 10^-10)) = 1 x 10^-5 M

So, the concentration of H3O+ is 1 x 10^-5 M, or 10 x 10^-6 M.