Section: Ionization expressions, Weak Bases
Using the equilibrium constants listed in your book, arrange the following .1 M aqueous solutions in order of increasing pH.
a)NaNO2
b)HCl
c)NaF
d)Zn(H2O)3(OH)(NO3)
Here's what I have so far:
a) NaNO2 --> Na+ + NO2-
NO2- + H2O <==> HNO2 + OH-
b) HCl + H2O <==> Cl- + H3O+
c) NaF --> F- + Na+
F- + H2O <==> HF + OH-
d) Zn(H2O)3(OH)(NO3) --> Zn(H2O)3(OH)+ + NO3-
NO3- + H2O <==> HNO3 + OH-
and I have the following constants available to me:
Ka:
HNO2= 6.0 x 10^-4
HF= 6.9 x 10^-4
Zn(H2O)4 +2= 3.3 x 10^-10
Kb:
NO2- = 1.7 x 10^-11
F- = 1.4 x 10^-11
Zn(H2O)3OH+ = 3.0 x 10^-5
I'm not sure what to do from here. Also, I'm thinking my equations for B and D are wrong, because none of the information/constants provided would help solve what I have now (also, B is Hydrochloric ACID, is it not?) Any help with this problem would be much appreciated.
Consider (a).
NO2- + H2O ==> HNO2 + OH-
Kb = (HNO2)(OH-)/(NO2-)
You start with 0.1 M NO2- and (HNO2)=(OH-)=0
When x mols NO2- reacts with H2O, x mols HNO2 and x mols OH- are formed.
At equilibrium, (HNO2) = x; (OH^-) = x; (NO2-) = 0.1-x.
Plug those into the Kb expression and solve for x = (OH^-), convert that to pOH and that to pH.
Do the same for all, then arrange in increasing order of pH.
HCl is a strong acid and has no Ka; that is why you can't find it. It is 100% ionized; therefore, (H^+) = 0.1 M and pH = ?
D is incorrect. NO3- is the anion of HNO3 which is a strong acid, like HCl, therefore, NO3- is not hydrolyzed. But look in your text for the hydrolysis of cations that are complex ions.
To solve this problem, you need to use the equilibrium expressions and the given equilibrium constants for each of the solutions. Let's go through each of the solutions one by one:
a) NaNO2:
The equation for the hydrolysis of NO2- is NO2- + H2O ⇌ HNO2 + OH-.
The given Kb for NO2- is 1.7 x 10^-11.
At equilibrium, let the concentration of OH- be x. Since NaNO2 is a strong electrolyte, the initial concentration of NO2- is 0.1 M.
Using the equilibrium expression Kb = (HNO2)(OH-)/(NO2-) and substituting the values, we get:
1.7 x 10^-11 = x * x / (0.1 - x).
Solve this equation for x, and you will get the concentration of OH-. Convert this concentration to pOH and then to pH using the equation pH + pOH = 14.
b) HCl:
HCl is a strong acid, so it is completely ionized in water. The concentration of H+ is 0.1 M, and you can calculate the pH directly using the equation pH = -log[H+].
c) NaF:
The equation for the hydrolysis of F- is F- + H2O ⇌ HF + OH-.
The given Kb for F- is 1.4 x 10^-11.
Follow the same process as in part (a) to solve for the concentration of OH- and then convert it to pH.
d) Zn(H2O)3(OH)(NO3):
Since NO3- is the conjugate base of a strong acid (HNO3), it does not undergo hydrolysis. However, Zn(H2O)3(OH)+ may undergo hydrolysis.
To determine whether Zn(H2O)3(OH)+ undergoes hydrolysis, you need to check if the Kb for Zn(H2O)3(OH)+ is given. If it is not given, you can assume it is negligible, and therefore, Zn(H2O)3(OH)+ will not affect the pH of the solution.
Once you have calculated the pH for each of the solutions, arrange them in increasing order of pH. The solution with the lowest pH will be the most acidic, and the solution with the highest pH will be the most basic.