The Kb for C5H5N (pyridine) is 1.7 x 10^-9. When pyridine reacts with water, it forms C5H5NH^+ and OH^- ions. To calculate the concentration of OH^- ions in solution, you would use the Kb value along with the initial concentration of pyridine and the equilibrium concentrations of pyridine, C5H5NH^+, and OH^- ions.
kb=[C5H5NH+][OH-]______[C5H5N]
The Kb for C5H5N (pyridine) is 1.7 x 10^-9. This value indicates the strength of the base in solution, with lower values suggesting a weaker base and higher values indicating a stronger base.
The Kb value for CH3NH2(aq) is 4.4 x 10^-4.
Kb=[(Ch3)3 NH+][OH-] __________ [(Ch3)3 N]
Kb=[HCN][OH-]/[CN-]
kb=[C5H5NH+][OH-]______[C5H5N]
The Kb for C5H5N (pyridine) is 1.7 x 10^-9. This value indicates the strength of the base in solution, with lower values suggesting a weaker base and higher values indicating a stronger base.
Kb = [CH3NH3 +] [OH-] / [CH3NH2]
The Kb value for CH3NH2(aq) is 4.4 x 10^-4.
Kb=[(Ch3)3 NH+][OH-] __________ [(Ch3)3 N]
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 Kb for (CH3)3N (trimethylamine) in water is a measure of the strength of the base (CH3)3NH in solution. It is used to calculate the equilibrium concentration of hydroxide ions (OH-) in solution when the base dissociates.
3.4 * 10^-8
KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l) NaOH(aq) + KHP(aq) => Na+ + KP- + H2O(l)
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) -> H2O(l) + Na+(aq) + Cl-(aq)
Na+ + OH- + H+ + Cl- H2O + Na+ + Cl-