How many moles of NaOH can be added to 1L solution 0.1M in NH3 and 0.1 M in NH4Cl without changing the POH by more than 1 unit kb NH3 = 1.8 into 10^_5
To solve this problem, we need to consider that NH3 is a weak base and NH4Cl is its conjugate acid. Given that the concentration of NH3 and NH4Cl in the solution is 0.1 M, we can use the Henderson-Hasselbalch equation to determine the pH of the solution, and then convert it to pOH.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant
[A-] = concentration of the conjugate base
[HA] = concentration of the acid
In this case, NH3 is the base (A-) and NH4Cl is the acid (HA). The pKa can be calculated using the pKa = -log10(Ka) equation. Since Kb + Ka = Kw (the ion product of water), we can use this relationship to find the pKa value.
Kw = Ka * Kb
Kw = 1.0 x 10^-14 (at 25 degrees Celsius)
Kb = 1.8 x 10^-5
Using the equation, we find:
Ka = Kw / Kb
Ka = 1.0 x 10^-14 / 1.8 x 10^-5
Ka ≈ 5.5 x 10^-10
Now that we have the pKa value, we can use the Henderson-Hasselbalch equation to find the pH of the solution:
pH = pKa + log([A-]/[HA])
We know that [A-] = [NH3] and [HA] = [NH4Cl] in this case. Let's assume that x moles of NaOH are added to the solution. Therefore, the concentration of NH3 will decrease by x moles and the concentration of NH4Cl will increase by x moles. We can now write the expression for the new concentrations:
[A-] = [NH3] - x
[HA] = [NH4Cl] + x
Using these new concentrations, we can substitute them into the Henderson-Hasselbalch equation:
pH = pKa + log(([NH3] - x) / ([NH4Cl] + x))
Since we want to find the change in pOH (ΔpOH), we need to convert the pH to pOH:
pOH = 14 - pH
To ensure that the pOH does not change by more than 1 unit, we can set up the following inequality:
|ΔpOH| ≤ 1
Now, let's solve this inequality.