what is the pH of a 0.10 M NH3???
Write the equilibrium equation.
NH3 + HOH ==> NH4^+ + OH^-
Now write the Keq expression.
I would use the ICE method and solve for (OH^-).
Post your work if you get stuck. You must look up Kb for NH3.
To determine the pH of a solution, we need to know the dissociation constant (Kw) of the solute and the concentration of the solute. In the case of ammonia (NH3), it acts as a weak base, so we need to know its dissociation constant (Kb) and its concentration.
The dissociation constant (Kb) for ammonia (NH3) is 1.8 x 10^-5. It represents the tendency of NH3 to accept a proton (H+) and form NH4+.
Given that the concentration of NH3 is 0.10 M, we can use the equation for the base dissociation constant to find the concentration of OH- ions (hydroxide ions), which will allow us to calculate the pOH. Then, we can use the relationship between pH and pOH to find the pH.
The equation for the base dissociation constant (Kb) is:
Kb = [NH4+][OH-] / [NH3]
Since we have the concentration of NH3 (0.10 M) and Kb, we can rearrange the equation to solve for [OH-].
Kb = [x][x] / (0.10 - x)
As ammonia (NH3) is a weak base, it undergoes partial dissociation. We can assume that the concentration of NH3 (0.10 M) remains relatively unchanged when compared to the extent of its dissociation, so we can approximate that (0.10 - x) is approximately equal to 0.10.
Now, we can rewrite the equation as:
Kb = [x][x] / 0.10
Simplifying further:
1.8 x 10^-5 = x^2 / 0.10
Rearranging the equation to solve for x (concentration of OH-):
x^2 = 1.8 x 10^-5 * 0.10
x^2 = 1.8 x 10^-6
Taking the square root of both sides:
x ≈ 4.24 x 10^-3 M
We now have the concentration of OH- ions (hydroxide ions). Next, we can calculate the pOH using the formula:
pOH = -log [OH-]
pOH = -log (4.24 x 10^-3) ≈ 2.37
Finally, we can find the pH using the relationship between pH and pOH:
pH + pOH = 14
pH = 14 - 2.37 ≈ 11.63
Therefore, the pH of a 0.10 M NH3 solution is approximately 11.63.