Wiki User
∙ 16y agoHCN + H20 => CN- + H3O+
0.45 X X
Solve for X
(4.0*10^-10) = x^2/0.45M
x=1.34e-5= [OH]
Find pOH= -log of the above number
then 14-4.87=pH= 9.13
Wiki User
∙ 16y agoIn a dynamic equilibrium with hydrogen cyanide dissolved in water, the majority of the solution will contain undissociated hydrogen cyanide molecules. This means that hydrogen cyanide (HCN) in its molecular form will be present in the highest concentration. There will also be some hydronium ions (H3O+) and cyanide ions (CN-) in solution due to partial dissociation.
The Kb for CN- (aq) is the equilibrium constant for the reaction of CN- with water to form HCN (aq) and OH- (aq). It represents the strength of the base CN- in solution. It can be calculated by taking the concentration of the products (HCN and OH-) and dividing by the concentration of CN- at equilibrium.
To find the Ka of HCN, first, calculate the concentration of H+ ions in the solution using the pH. Then, since HCN is a weak acid, assume that [H+] = [CN-]. Use the equilibrium expression for HCN dissociation, HCN ⇌ H+ + CN-, and the initial concentration of HCN to calculate the Ka value.
HCN is a weak acid because it does not completely dissociate in water to release all of its H+ ions. Instead, only a small fraction of HCN molecules dissociate into H+ and CN- ions, resulting in a low concentration of H+ ions in solution. This low concentration of H+ ions makes HCN a weak acid.
ka=[H+][CN-]/[HCN]
HCN is a linear molecule.
HCN has a linear molecule.
HCN is hydrocyanic acid and is acidic (not basic).
NaCN + H2O ---> CN-+ H3O++ Na+ In that equation Na+ is just a spectator ion,, further reaction with water results in: CN- + H2O ---> HCN + OH- thus causing the resulting equation to be basic
The conjugate base of HCN is CN-. It is formed when HCN donates a proton (H+) and becomes negatively charged.
The nomenclature for HCN is hydrogen cyanide.
HCN is an acid; KClO3 is a salt.