You have an unknown concentration of ammonia. When you add 15.0mL of a 0.400 M HCl solution to 50.0mL of the ammonia solution, the pH of the solution becomes 8.75. What is the concentration of the original ammonia solution? [Kb (NH3) = 1.8 x 10^-5]
Use the Henderson-Hasselbalch equation.
15.0 mL x 0.400M = 6.0 millimols HCl
50 mL x ?M NH3 = ??
...........NH3 + HCl ==> NH4^ + Cl^-
...........x.......0.......0
add.............6.0.........
change...-6.0..-6.0.......6.0
equil....x-6.0...0.........6.0
pH = pKa + log (base)/(acid)
8.75 = 9.26 + log(x-6)/(6)
Solve for x = millimols NH3
M NH3 = millimols/50 mL
.1132 M @ equilibrium
To find the concentration of the original ammonia solution, we can begin by applying the concept of acid and base reactions and the Kw expression.
1. Write the balanced chemical equation for the reaction between ammonia (NH3) and hydrochloric acid (HCl):
NH3 + HCl → NH4+ + Cl-
2. Since ammonia is a weak base and hydrochloric acid is a strong acid, we can assume that the ammonia reaction goes to completion, and the concentration of NH3 will be equal to the concentration of NH4+ formed:
[NH3] = [NH4+]
3. Use the concentration of HCl and the volume of the solution to calculate the number of moles of HCl added:
Moles of HCl = concentration of HCl (M) × volume of HCl (L)
Moles of HCl = 0.400 M × 0.015 L = 0.006 moles
4. Since the reaction between HCl and NH3 is a 1:1 ratio, the number of moles of NH4+ formed will be equal to the number of moles of HCl added.
5. Determine the number of moles of NH4+ by using the concentration of NH4+ (which is equal to the concentration of NH3 in this case) and the volume of the solution:
Moles of NH4+ = concentration of NH4+ (M) × volume of solution (L)
= concentration of NH3 (M) × volume of solution (L)
Moles of NH4+ = [NH3] × (0.050 L + 0.015 L) = [NH3] × 0.065 moles
6. Substitute the value obtained for the moles of NH4+ formed (which is equal to the moles of NH3) into the equilibrium expression for the Kb of NH3:
Kb = [NH4+][OH-] / [NH3]
Since we are looking for the concentration of the original ammonia solution, we can rewrite the expression as follows:
Kb = [NH3][OH-] / [NH3]
Kb = [OH-] (Since [NH3] cancels out)
7. Rearrange the equation to solve for [OH-]:
[OH-] = Kb = 1.8 × 10^-5 = 1.8 × 10^-5 M
8. Since the pH is given as 8.75, we know that pOH = 14 - pH:
pOH = 14 - 8.75 = 5.25
9. Convert pOH to [OH-] concentration by taking the antilogarithm:
[OH-] = 10^(-pOH)
[OH-] = 10^(-5.25) = 5.6 × 10^(-6) M
10. Since [OH-] is equal to the concentration of NH3, we have found the concentration of the original ammonia solution:
[NH3] = 5.6 × 10^(-6) M
Therefore, the concentration of the original ammonia solution is 5.6 × 10^(-6) M.