calculate the pH of a 0.2moldm-3 solution of ammonia at 298K given that Kb for ammonia is 1.8x10-5
To calculate the pH of a solution of ammonia, you need to consider the equilibrium reaction of ammonia with water:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant for this reaction, known as the base dissociation constant (Kb), is given as 1.8x10-5.
The Kb expression for this reaction is:
Kb = [NH4+][OH-] / [NH3]
In this case, we know the concentration of ammonia ([NH3]) is 0.2 moldm-3. Let's assume that at equilibrium, x mol dm-3 of ammonia reacts to form x mol dm-3 of NH4+ and OH-.
Using this information, we can write the expression for Kb in terms of x:
1.8x10-5 = (x)(x) / (0.2 - x)
At this point, we can make an assumption to simplify the calculation. Since Kb is very small, we can approximate (0.2 - x) to be 0.2. This approximation is valid as long as the value of x is much smaller than 0.2. We will verify this assumption later.
Now, we can simplify the expression:
1.8x10-5 = (x)(x) / 0.2
Rearranging the equation, we get:
x^2 = (1.8x10-5)(0.2)
x^2 = 3.6x10-6
Taking the square root of both sides:
x ≈ 1.9x10-3 moldm-3
Now, let's check if our assumption was valid. Since x is indeed much smaller than 0.2, our approximation holds.
Next, we can calculate the concentration of OH- ions:
[OH-] = x = 1.9x10-3 moldm-3
Since water is neutral, the concentration of H+ ions is equal to the concentration of OH- ions:
[H+] = [OH-] = 1.9x10-3 moldm-3
Finally, we can calculate the pOH:
pOH = -log[OH-] = -log(1.9x10-3) = 2.72
Since pH + pOH = 14 (at 298K), we can find the pH:
pH = 14 - pOH = 14 - 2.72 = 11.28
Therefore, the pH of the 0.2 moldm-3 solution of ammonia at 298K is approximately 11.28.