Question

3. Calculate the equilibrium constant for the reaction: CN- + H2O == HCN + OH- A.)5.0*10^-4 B.) 5.0*10^-5 C.) 2.0*10^-5 D.) 2.0*10^-4 E.) 4.9*10^-10

Ka Values: HF, 6.8*10^-4 HNO2, 4.5*10^-4 HOBr, 2.5*10^-9 NH4+, 5.6*10^-10 HCN, 4.9*10^-10

Answer 1

CN^- + H₂O <===> HCN + OH^- Ka = for HCN = 4.9 x 10^-10

he has given Ka of HCN <-----> H+ + CN- Ka = 4.9 x 10^-10 ----1

but here water is also getting dissociating

H2O <====> H+ + OH-   Ka for water = 1 x 10¹⁴ ----2

reaction 1 and two together is final reaction

K = 4.9 x 10^-10 x 1 x 10¹⁴

= 4.9 x 10⁴

^{wit hole reaction is in reverse direction}

K = 1/4.9 x 10⁴

= 2.0 x 10^-5

so answer is C