AP Chem pH question?

Question

How would I calculate the pH of a 0.20M KNO2 sol*n? Same for a 0.37M NH4ClO4 sol*n? Any help is appreciated!

Answer 1

A 0.20 M KNO2 solution contains K+ ions and NO2- ions, along with water. In this case, you only are concerned with the NO2- ions because K+ will be negligible where the pH is concerned. NO2- is the conjugate base of a weak acid (HNO2) so it will act as a base. Write the reaction that shows NO2- acting as a base.

R: NO2- + H2O <--> HNO2 + OH-
I:...0.20......................0...........0
C:..-x.........................+x........+x
E: 0.20-x....................x...........x
(HNO2 and OH- are both initially zero, have a change value of +x and an equilibrium concentration of x)
The equilibrium expression for this reaction is:
Kb=[HNO2][OH-]
....----------------------
..........[NO2-]
With the Ka for HNO2, you can solve for the Kb of NO2-. Ka for HNO2 is 4.6 x 10^-4
Ka x Kb = Kw(1.0 x 10^-14)
Solving for the Kb, you get 2.2 x 10^-11

Plug in the equilibrium values and the calculated Kb, and solve for x, which will end up being your concentrations of HNO2 and OH-
2.2 x 10^-11=[x][x]
...................-------------
.....................[.2-x]
Solving for the quadratic equation, you get x=2.1 x 10^-6=[OH-] and [HNO2]

OH- is a strong base and HNO2 is a weak acid, so the pH will be governed by the strong base.
pOH= -log[OH-]
-log[2.1 x 10^-6]
pOH=5.68
pH+pOH=14
pH=14-5.68
pH=8.32

For the other solution, you would proceed similarily, except in this case you would have NH4+ and ClO4-. ClO4- is considered negligible, and NH4+ is the conjugate acid of a weak base (NH3). Show NH4+ acting as an acid in water
R: NH4+ + H2O <--> NH3 + H3O+
The set up is the same except the equilibrium concentration of NH4+ is .37-x

You will need to find the Ka for NH4+ from the Kb of NH3. Kb for NH3 is 1.8 x 10^-5, so from the equation above, Ka= 5.6 x 10^-10.
Ka=[NH3][H3O+]
.....---------------------
..........[NH4+]

5.6 x 10^-10=[x][x]
...............---------------
..................[.37-x]

In this case you can assume x to be negligible where the equilibrium concentration of NH4+ is concerned, so [NH4+] at equilibrium =.37
The equation is now:
x^2=2.072 x 10^-10
x=1.4 x 10^-5
This is the concentration of a weak base NH3, and the hydronium ion, which is a strong acid, which will govern the pH in this case.

pH=-log[H+]
-log[1.4 x 10^-5]
pH=4.85

Hope this helps! Sorry if I made any mistakes

Answer 2

Take the pH of the .20M and raise the power of .37M

Answer 3

use an ICE table