Determine the pH of 263 ml of solution which has [NH 4I] = 0.300 M. Kb = 1.74 × 10-5 for NH3(aq)
I managed to get it :)
though for others who might be looking to solve this for their practice questions
Convert Kb to Ka
Ka = kw/Kb
then just use an ice table with Ka=x^2/ 0.300M-x (x is small)
and just -log[H3O+)
4.9
To determine the pH of a solution with [NH4I] = 0.300 M, we need to consider the NH3 formed by the reaction of NH4I with water.
The balanced chemical equation for the reaction is:
NH4I (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
In this reaction, NH4I acts as an acid and donates a proton (H+) to water to form NH3 and H3O+.
First, we need to calculate the concentration of NH3 produced. Since there is a one-to-one ratio between NH4I and NH3, the concentration of NH3 will also be 0.300 M.
Next, we need to calculate the concentration of H3O+ (hydroxonium ion) using the Kb of NH3. Kb is the equilibrium constant for the reaction of NH3 with water to form OH- and NH4+. However, since we are given the Kb value, we can use it to find the concentration of H3O+.
Kb = [NH4+][OH-] / [NH3]
We know that [NH3] = 0.300 M and Kb = 1.74 × 10^(-5). Let's assume x is the concentration of OH-.
1.74 × 10^(-5) = (x)(0.300) / (0.300)
Simplifying the equation, we get:
1.74 × 10^(-5) = x
Now we know that the concentration of H3O+ = x. Since water is neutral, [H3O+] = [OH-], and therefore the concentration of OH- is also x.
So, the concentration of H3O+ or [H+] is 1.74 × 10^(-5) M.
To find the pH, we need to take the negative logarithm of [H3O+] (or [H+]):
pH = -log10 (1.74 × 10^(-5))
Calculating this expression, we find:
pH ≈ 4.76
Therefore, the pH of the solution is approximately 4.76.