HCN + 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
C2H3O2-(aq) H+(aq) H3O+(aq) OH-(aq) HCN(aq)Those are the choices that could be given, but the answer is HCN. It can't be H+ or OH- because the water is de-ionized and H3O+ is not possible.
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.
A pH of 4.29 corresponds to a hydrogen ion concentration of 10-4.29, which equals 10-5 X 10+0.71, since 0.71 - 5 = -4.29. 10+0.71 = 5.1, to the justified number of significant digits. By definition, Ka = [H+]X[CN-]/[HCN], and in the absence of other sources of cyanide anions, [H+] = [CN-]. Within the number of significant digits given, [HCN] = 0.16, and Ka then = (5.1 X 10-5)2/0.16 = 1.6 X 10-8.
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).
The conjugate base of HCN is CN-. It is formed when HCN donates a proton (H+) and becomes negatively charged.
HCN is an acid; KClO3 is a salt.
hydrogen cyanide
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