ka=[H+][NO2_]/[HNO2]
HCN(aq) ==> H^+(aq) + CN^-(aq)Ka = [H+][CN-]/[HCN] and the value can be looked up in a book or on line.
Na+OH-(aq) + H+Cl-(aq) Na+Cl-(aq) + H+2O-(l)
HCLO4(aq)+H2O(L) → H3O+(aq)+CLO4-(aq) Or HCLO4(aq) → H++CLO4-
say for example the equilibrium I2(aq)+H2O(l)-----HOI(aq)+I(aq)+H(aq) Think Lechatlier principle... addition of NaOH will cause the H ions to react with the OH ions to cause more water (more reactants) increase in reactants shifts the equilibrium in the FORWARD direction to form MORE H+ to restore the equilibrium
not sure
Ka= [H+][NO2-] [HNO2]
ka=[H+][NO2_]/[HNO2]
ka=[H+][CN-]/[HCN]
Ka= [H+] [H2BO3-] / [h3BO3] (Apex)
Ka= [h+][HCO3-]/[H2CO3]
ka=[H+][NO2_]/[HNO2]
HCN(aq) ==> H^+(aq) + CN^-(aq)Ka = [H+][CN-]/[HCN] and the value can be looked up in a book or on line.
Since H3PO4 has 3 ionizable hydrogens, it will have three Ka values. Approximate values areKa1 = 7x10^-3; Ka2 = 6x10^-8 and Ka3 = 4.5x10^-13
HCl (aq) + H2O (L) ---------> H3O+ (aq) + Cl- (aq)
H+(aq)+ Br-(aq) +Na+(aq) + OH-(aq)----->H+(l)+OH-(l) +Br-(S) +Na+(S)By cancelling out the spectator ions we are left with...H+(aq) + OH-(aq) ------> H(l)+ + OH(l)-
Molecular Eq HC2H3O2(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq) Ionic Eq H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq) Net Ionic Eq H+(aq) + NH3(aq) -> NH4+(aq)