What is the ph of a soulution made by adding 0.20 mole NH3(aq) to 0.30 mole NH4Cl and 0.10 mole KOH to 100ml of water?
Kb for NH3 (aq)=1.8x10^-5M
I would do this.
.......NH4^+ + OH^- ==> NH3 + H2O
I......0.3......0.......0.2.......
add............0.1................
C,....0.3-0.1..-0.1.....+0.1
E......0.2......0........0.3
Convert the mols in the E line to M (mols/L) and substitute into the Henderson-Hasselbalch equation.
To find the pH of the solution, we need to determine the concentration of OH- ions in the solution. Given that 0.10 mole of KOH reacts with water, we can first calculate the concentration of OH- ions contributed by KOH.
1. Calculate the molarity (M) of KOH:
Molarity (KOH) = moles of KOH / volume of solution in liters
Since the volume of the solution is given as 100 ml (0.100 L) of water, the molarity of KOH is:
Molarity (KOH) = 0.10 moles / 0.100 L = 1.00 M
2. Using the fact that KOH is a strong base, we know that it dissociates completely, so the concentration of OH- ions contributed by KOH is also 1.00 M.
Next, we need to determine the concentration of OH- ions contributed by the reaction between NH3 (aq) and water.
3. NH3 (ammonia) is a weak base, so we need to use its Kb value to calculate the concentration of OH- ions.
Kb = [OH-] * [NH4+] / [NH3]
Since NH4Cl reacts with NH3 to form NH4+ ions, we can calculate the concentration of NH4+ using the moles of NH4Cl and the volume of the solution in liters:
Molarity (NH4+) = moles of NH4Cl / volume of solution in liters
Molarity (NH4+) = 0.30 moles / 0.100 L = 3.00 M
4. Using the balanced equation for the reaction between NH3 and water, we know that one molecule of NH3 reacts with one molecule of water to produce one NH4+ ion and one OH- ion. This means that the concentration of OH- ions is equal to the concentration of NH4+ ions:
[OH-] = [NH4+] = 3.00 M
Now we can calculate the total concentration of OH- ions in the solution:
Total [OH-] = [OH-] (from KOH) + [OH-] (from NH4+)
= 1.00 M + 3.00 M
= 4.00 M
The pH of a solution can be calculated using the formula:
pOH = -log [OH-]
5. Calculate the pOH of the solution:
pOH = -log (4.00 M) = -log (4.00) = 0.398
6. Finally, we can calculate the pH of the solution using the relation:
pH + pOH = 14
pH = 14 - pOH
= 14 - 0.398
= 13.602
Therefore, the pH of the solution is approximately 13.602.