For each solute listed, write the chemical equation for the dissociation of the
ID: 825298 • Letter: F
Question
For each solute listed, write the chemical equation for the dissociation of the salt into its ions. If the salt is acidic, write the chemical equation for the reaction of the acidic ion in water to produce hydronium ion, write the expression for the acid ionization constant, Ka, and calculate the value of Ka.
If the salt is basic, write the chemical equation for the reaction of the basic ion with water to produce hydroxide ion, write the expression for the base ionization constant, Kb, and calculate the value of Kb from your data.
Solute
Chemical equations
Expression for Kb or Ka
Value of Kb or Ka
NaC2H3O2
Na2SO4
NaHSO4
Na2CO3
NH4Cl
Solute
Chemical equations
Expression for Kb or Ka
Value of Kb or Ka
NaC2H3O2
Na2SO4
NaHSO4
Na2CO3
NH4Cl
Explanation / Answer
NaC2H3O2 --> Na + C2H3O2-
then C2H3O2- + H2O <--> HC2H3O2 + OH-
Kb = [HC2H3O2 ][OH-]/[C2H3O2]
Kb = Kw/Ka = (10-14)/1.8*10^-5) = 5.55*10^-10
Na2SO4
2NA +´SO4-2
SO4-2 + H2O <--> HSO4- + OH-
Kb = [HSO4-][OH-]/[SO4-2]
Kb = Kw/Ka = (10^-14)/(1.2*10^-2) = 8.33*10^-13
NaHSO4
NaHSO4 --> Na+ + HSO4-
HSO4- <--> H+ + SO4-2
Ka = [H+][SO4-2]/[HSO4-]
1.2*10^-2 = [H+][SO4-2]/[HSO4-]
Na2CO3
Na2cO3 --< 2Na+ + CO3-2
CO3-2 + H2O <--> HCO3- + OH-
Kb2 = Kw/Ka2 = (10^-14)/(4.8*10^-11) = 2.08*10^-4
NH4Cl
NH4+ + Cl-
NH4+ --> NH3 + H+
Ka = [NH3][H+]/[NH4+]
Ka = Kw/Kb = (10^-14)/(1.8*10^-5 = 5.55*10^-10
+
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