The equation you provided appears to represent a chemical reaction involving cyanide ions (CN⁻) and water (H₂O), resulting in the formation of hydrogen cyanide (HCN) and hydroxide ions (OH⁻). In this process, the cyanide ion interacts with water, leading to the release of HCN and the production of hydroxide. This reaction can be understood as a hydrolysis of the cyanide ion, where it accepts a proton from water.
The acid in the reaction is hydrogen cyanide (HCN), which is formed when cyanide ion (CN-) reacts with water (H2O) to release hydroxide ion (OH-).
Kb = [CH3NH3 +] [OH-] / [CH3NH2]
The reaction is:MgO + H2O = Mg(OH)2
Ch3coo(-) + h2o ---> ch3cooh + oh(-)
The concentration of OH- decreases as the concentration of H+ increases. This is beacause there is an equilibrium H2O <-> H+ + OH- and therefore the [H+][OH-] is a constant
The acid in the reaction is hydrogen cyanide (HCN), which is formed when cyanide ion (CN-) reacts with water (H2O) to release hydroxide ion (OH-).
Kb=[HCN][OH-]/[CN-]
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.
The eqnet ionic equation is HCN + OH- --> H2O + CN-
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
NH4OH + HC2H3O2 ---> NH4C2H3O2 + H2ONH4+ + OH- + H+ + C2H3O2- ---> NH4+ + C2H3O2- + H2OOH- (aq) + H+ (aq)---> H2O (l)
NH3 + H20 <----> NH4+ + OH- Ammonia is a weak base so it is the favored side of the equilbrium. Conjugate acid and base pairs only differ by a proton. So ammonia and ammonium are pairs and water and hydroxide ions are pairs. NH4+ + CN- <-------> HCN + NH3
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
The net ionic equation for the given reaction is H+ (aq) + OH- (aq) → H2O (l)
Sodium cyanide is a base/salt that dissociates in water. CN- is a conjugate base of a weak acid so it grabs a proton (in small amounts) from the water molecule to become HCN.
The balanced equation for MgO + H2O is MgO + H2O -> Mg(OH)2.
Kb = [CH3NH3 +] [OH-] / [CH3NH2]