calculate the pH of 0.0125M NH3 Kb=1.8x10-5
To calculate the pH of a 0.0125M NH3 solution, we first need to calculate the concentration of OH- ions resulting from the dissociation of NH3 in water.
NH3 + H2O ⇌ NH4+ + OH-
We can use the Kb expression to set up an equation for the reaction:
Kb = [NH4+][OH-] / [NH3]
We know the initial concentration of NH3 is 0.0125M. Let's assume x moles of NH3 react to form x moles of NH4+ and OH- ions:
Kb = [x][x] / [0.0125 - x]
Given that Kb = 1.8 x 10^(-5), we can set up the equation:
1.8 x 10^(-5) = x^2 / [0.0125 - x]
Since Kb is very small, we can assume that x is also very small, so [0.0125 - x] is approximately 0.0125. Then:
x^2 ≈ (1.8 x 10^(-5))(0.0125)
x^2 ≈ 2.25 x 10^(-7)
x ≈ √(2.25 x 10^(-7))
x ≈ 1.5 x 10^(-4)
Now we have determined that the concentration of OH- ions is approximately 1.5 x 10^(-4) M. To find the pH, we will first calculate the pOH using the following formula:
pOH = -log[OH-]
pOH = -log (1.5 x 10^(-4))
pOH ≈ 3.82
Finally, we can determine the pH using the relationship between pH and pOH:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 3.82
pH ≈ 10.18
So the pH of the 0.0125M NH3 solution is approximately 10.18.