Calculate the pH for each of the following solutions:
(a) 0.10 M NH3
(b) 0.050 M C5H5N (pyridine)
To calculate the pH for each of the solutions, we need to use the relationship between the concentration of hydrogen ions (H+) and hydroxide ions (OH-) in an aqueous solution.
(a) For 0.10 M NH3:
NH3 is the chemical formula for ammonia, which acts as a weak base. In water, it reacts with water to produce the hydroxide ion (OH-) and ammonium ion (NH4+), according to the following equilibrium equation:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant expression (Kw) for this reaction is:
Kw = [NH4+][OH-] / [NH3]
Since we have excess NH3, we can assume that the concentration of NH3 remains constant and neglect it in the equation. Therefore, the reaction can be simplified to:
NH3 + H2O ⇌ NH4+ + OH-
In this case, the concentration of NH3 will be equal to the concentration of NH4+. Therefore, [NH4+] = 0.10 M. Since we know [NH4+], we need to find [OH-] to calculate the pH. To do this, we use the fact that water is neutral and Kw = [H+][OH-], where Kw = 1.0 x 10^-14 at 25°C.
[H+][OH-] = 1.0 x 10^-14
[NH4+][OH-] = 1.0 x 10^-14 (NH4+ = [NH3])
0.10[OH-] = 1.0 x 10^-14
[OH-] = (1.0 x 10^-14) / 0.10
[OH-] = 1.0 x 10^-13 (approximate)
Now, we can find the pOH:
pOH = -log[OH-]
pOH = -log(1.0 x 10^-13)
pOH ≈ 13
Since pH + pOH = 14, we can calculate the pH as:
pH = 14 - pOH
pH = 14 - 13
pH ≈ 1
Therefore, the pH for a 0.10 M NH3 solution is approximately 1.
(b) For 0.050 M C5H5N (pyridine):
C5H5N is the chemical formula for pyridine, which acts as a weak base. Similar to the previous calculation, we need to consider the equilibrium between pyridine and water:
C5H5N + H2O ⇌ C5H5NH+ + OH-
Again, we can assume that the concentration of pyridine remains constant and neglect it in the equation. Therefore, the reaction can be simplified to:
OH- = [C5H5NH+]
To calculate [OH-], we need [C5H5NH+]:
[C5H5NH+] = 0.050 M
Since [C5H5NH+] = [OH-], we have:
[OH-] = 0.050 M
We can find the pOH:
pOH = -log[OH-]
pOH = -log(0.050)
pOH ≈ 1.3
Again, using the relationship pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH
pH = 14 - 1.3
pH ≈ 12.7
Therefore, the pH for a 0.050 M C5H5N (pyridine) solution is approximately 12.7.
To calculate the pH of a solution, we need to determine the concentration of hydrogen ions (H+) in the solution. The concentration of H+ ions is related to the concentration of hydroxide ions (OH-) by the equation:
Kw = [H+][OH-]
Where Kw is the ion product constant for water, which is equal to 1.0 x 10^-14 at 25°C.
For weak acids or bases, we use the equilibrium expression for the ionization reaction to determine the concentration of H+ or OH- ions.
(a) 0.10 M NH3:
NH3 is a weak base. It reacts with water to form the ammonium ion (NH4+) and hydroxide ion (OH-).
The ionization reaction is:
NH3 + H2O ⇌ NH4+ + OH-
Since NH3 is a weak base, we can assume that the concentration of NH3 is essentially equal to the concentration of OH-. Therefore, we can write:
Kw = [NH4+][OH-]
1.0 x 10^-14 = (0.10)(x)
Solving for x, we get:
x = [OH-] = 1.0 x 10^-13 M
Since [OH-] = [NH3], the concentration of NH3 is also 1.0 x 10^-13 M.
The concentration of H+ is given by:
[H+] = Kw / [OH-]
= 1.0 x 10^-14 / 1.0 x 10^-13
= 1.0 x 10^-1 M
To calculate the pH, we use the equation:
pH = -log[H+]
pH = -log(1.0 x 10^-1)
pH = 1.0
Therefore, the pH of a 0.10 M NH3 solution is 1.0.
(b) 0.050 M C5H5N (pyridine):
C5H5N is a weak base. It reacts with water to form the pyridinium ion (C5H5NH+) and hydroxide ion (OH-).
The ionization reaction is:
C5H5N + H2O ⇌ C5H5NH+ + OH-
Similarly to part (a), we can assume that the concentration of pyridine (C5H5N) is essentially equal to the concentration of OH-, so we write:
Kw = [C5H5NH+][OH-]
1.0 x 10^-14 = (0.050)(x)
Solving for x, we get:
x = [OH-] = 2.0 x 10^-13 M
Since [OH-] = [C5H5N], the concentration of C5H5N is also 2.0 x 10^-13 M.
The concentration of H+ is given by:
[H+] = Kw / [OH-]
= 1.0 x 10^-14 / 2.0 x 10^-13
= 5.0 x 10^-2 M
To calculate the pH, we use the equation:
pH = -log[H+]
pH = -log(5.0 x 10^-2)
pH = 1.3
Therefore, the pH of a 0.050 M C5H5N (pyridine) solution is approximately 1.3.
NH3 + HOH ==> NH4 + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
Prepare an ICE chart, plug into the Ka expression, and solve for (OH^-). Convert to pOH, then to pH.
The pyridine is worked the same way.
C5H5N + HOH ==> C5H5NH^+ + OH^-
Post your work if you get stuck.