HCN(aq) ==> H^+(aq) + CN^-(aq)Ka = [H+][CN-]/[HCN] and the value can be looked up in a book or on line.
ka=[H+][NO2_]/[HNO2]
Kb = [CH3NH3 +] [OH-] / [CH3NH2]
The cell notation for the given redox reaction is: Pb₂⁺(aq) | Pb(s) || Cr(s) | Cr³⁺(aq)
A precipitate
HCN(aq) ==> H^+(aq) + CN^-(aq)Ka = [H+][CN-]/[HCN] and the value can be looked up in a book or on line.
3.4 * 10^-8
Ka= [H+] [H2BO3-] / [h3BO3] (Apex)
ka=[H+][NO2_]/[HNO2]
ka=[H+][CN-]/[HCN]
6.6 x 10-9
Ka= [h+][HCO3-]/[H2CO3]
The equilibrium constant expression for the reaction you provided would be ( K_a = \frac{[H^+][H_2BO_3^-]}{[H_3BO_3]} ). However, the specific value of ( K_a ) for this reaction would depend on the concentrations of the species involved in the particular experimental conditions.
The Ka for the dissociation of HNO2 is 4.5 x 10^-4. This value represents the equilibrium constant for the reaction of HNO2 dissociating into H+ and NO2-.
not sure
Ka= [H+][NO2-] [HNO2]
Kb = [CH3NH3 +] [OH-] / [CH3NH2]