Asked by Ahmed
Find Net Ionic equation for hydrolysis , Expression for equilibrium constant (Ka or Kb) and Value of (Ka or Kb)
Net Ionic equations I've got
NaC2H3O2 == CH3COO^-+H2O -->CH3COOH+OH^-
Na2CO3 ==== CO3 + 2H2O → H2CO3 + 2-OH
Kb = [H2CO3] [-OH]2 / [CO3].
Kb = 1/Ka1Ka2 = 7.1x10^13
NH4Cl === NH4Cl+H2O --> NH4OH + HCl
ZnCl2 === Zn2+H2O--> ZnO+2H^+
KAl(SO4)2 === Al^+3 + 3OH--> Al(OH)3
I have no idea about the rest.
Thank you!!!
Net Ionic equations I've got
NaC2H3O2 == CH3COO^-+H2O -->CH3COOH+OH^-
Na2CO3 ==== CO3 + 2H2O → H2CO3 + 2-OH
Kb = [H2CO3] [-OH]2 / [CO3].
Kb = 1/Ka1Ka2 = 7.1x10^13
NH4Cl === NH4Cl+H2O --> NH4OH + HCl
ZnCl2 === Zn2+H2O--> ZnO+2H^+
KAl(SO4)2 === Al^+3 + 3OH--> Al(OH)3
I have no idea about the rest.
Thank you!!!
Answers
Answered by
DrBob222
#1 you have is close but not right.
I would have written for the hydrolysis of NaC2H3O2 as a net ionic equation.
CH3COO^- + H2O ==> CH3COOH + OH^-
Kb for CH3COO^- = (Kw/Ka for CH3COOH) = about 1E-14/1.8E-5 = ?
CO3^2- + HOH ==> HCO3^- + OH^-
Kb for CO3^2- = (Kw/K2 for H2CO3) = ?
NH4^+ + H2O ==> H3O^+ + NH3
Ka for NH4^+ = (Kw/Kb for NH3) = ?
Al^3+ + 6HOH ==> Al(H2O)6^3+
Then Al(H2O)6^3+ + H2O ==> Al(H2O)5^2+ + H3O^+
Ka = Kw/Kb
I would have written for the hydrolysis of NaC2H3O2 as a net ionic equation.
CH3COO^- + H2O ==> CH3COOH + OH^-
Kb for CH3COO^- = (Kw/Ka for CH3COOH) = about 1E-14/1.8E-5 = ?
CO3^2- + HOH ==> HCO3^- + OH^-
Kb for CO3^2- = (Kw/K2 for H2CO3) = ?
NH4^+ + H2O ==> H3O^+ + NH3
Ka for NH4^+ = (Kw/Kb for NH3) = ?
Al^3+ + 6HOH ==> Al(H2O)6^3+
Then Al(H2O)6^3+ + H2O ==> Al(H2O)5^2+ + H3O^+
Ka = Kw/Kb
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