Determine the pH of 263 ml of solution which has [NH 4I] = 0.300 M. Kb = 1.74 × 10-5 for NH3(aq)
To determine the pH of the solution, we need to calculate the concentration of OH- ions in the solution and then convert it to pH.
The given solution contains NH4I, which dissociates in water as follows:
NH4I(aq) ⇌ NH4+(aq) + I-(aq)
Since NH4I dissociates completely, the initial concentration of NH4+ ions in the solution is equal to the concentration of NH4I, which is 0.300 M.
To find the concentration of OH- ions, we need to determine the concentration of NH3(aq) formed when NH4+ ions react with water. This can be done by calculating the concentration of NH4+ ions that react with OH- ions from water.
The reaction between NH4+ ions and OH- ions can be represented as follows:
NH4+(aq) + OH-(aq) ⇌ NH3(aq) + H2O(l)
Since NH4+ ions react with OH- ions to form NH3(aq), the concentration of NH3(aq) formed would be equal to the concentration of NH4+ ions reacting.
Let's assume x is the concentration of NH3(aq) formed and the concentration of OH- ions produced.
From the equation, we can see that the concentration of OH- ions is equal to the concentration of NH4+ ions reacting, which is 0.300 M.
Now, we can set up an equation using the Kb value to find the concentration of OH- ions:
Kb = [NH3][OH-] / [NH4+]
Substituting the known values:
1.74 × 10-5 = x * x / (0.300 - x)
Since x is small compared to 0.300 M, we can approximate 0.300 - x to be approximately 0.300 M.
1.74 × 10-5 = x * x / 0.300
Rearranging the equation:
x^2 = 0.300 * 1.74 × 10-5
x^2 = 5.22 × 10-6
Taking the square root of both sides:
x ≈ 2.29 × 10-3 M
The concentration of OH- ions is approximately 2.29 × 10-3 M.
Now, to find the pH, we can use the equation:
pOH = -log[OH-]
pOH = -log(2.29 × 10-3)
pOH = -(-2.638) (using a calculator)
pOH ≈ 2.638
Finally, we can find the pH using the equation:
pH = 14 - pOH
pH = 14 - 2.638
pH ≈ 11.36
Therefore, the pH of the solution is approximately 11.36.
To determine the pH of the solution, we need to find the concentration of hydroxide ions (OH-) and then convert it to pH.
First, we need to find the concentration of ammonia (NH3) in the solution. Since ammonia is a weak base, we can use the given Kb value to find the concentration of hydroxide ions.
The Kb expression for the reaction of ammonia with water is:
Kb = [NH4+][OH-] / [NH3]
Assuming the initial concentration of OH- is x, the concentration of NH4+ is also x since the starting concentration of ammonia is zero, and the concentration of NH3 is 0.300 M, the equation becomes:
1.74 × 10-5 = (x)(x) / 0.300
Simplifying the equation:
x^2 = 1.74 × 10-5 * 0.300
x^2 = 5.22 × 10-6
Taking the square root of both sides, we find:
x = √(5.22 × 10-6)
x ≈ 0.00228 M
Now, we need to find the pOH of the solution using the concentration of OH-:
pOH = -log[OH-]
pOH = -log(0.00228)
pOH ≈ 2.64
Finally, we can find the pH of the solution using the pOH:
pH = 14 - pOH
pH = 14 - 2.64
pH ≈ 11.36
So, the pH of the solution is approximately 11.36.
NH4+ + H2O ==> H3O+ + NH3
Ka = (Kw/Kb) = (H3O+)(NH3)/(NH4+)
Set up an ICE chart and substitute into the Ka expression. Solve for (H3O+) and convert to pH. You don't need the 263 mL.