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An aqueous solution contains 0.469 m ammonia. Calculate the ph of the solution after the addition of 1.83×10^-2 moles of hydroiodic acid to 155 ml of this solution. The volume does not change.
Accepted Answer
Step 1: Calculate the moles of ammonia present in 155 mL of the initial solution.
Moles of ammonia = 0.469 M * 0.155 L = 0.0726 mol
Step 2: Calculate the concentration of hydroiodic acid added to the solution.
[HI] = 1.83 × 10^-2 mol / 0.155 L = 0.118 M
Step 3: Set up an ICE table to determine the equilibrium concentrations of all species.
| | Initial | Change | Equilibrium |
|:----------|:----------:|:----------:|:----------:|
| NH3 | 0.469 M | -x | (0.469 - x) M |
| HI | 0.118 M | -x | (0.118 - x) M |
| H3O+ | 0 | +x | x M |
| NH4+ | 0 | +x | x M |
Step 4: Write the equilibrium constant expression for the dissociation of hydroiodic acid.
Ka = [H3O+][I-] / [HI]
Step 5: From the ICE table and the Ka expression, set up an equation to solve for x.
Ka = x^2 / (0.118 - x)
10^-10 = x^2 / (0.118 - x)
Step 6: Solve for x.
x = 1.05 × 10^-6 M
Step 7: Calculate the pH of the solution.
pH = -log[H3O+]
= -log(1.05 × 10^-6)
= 5.98
Therefore, the pH of the solution after the addition of hydroiodic acid is 5.98.
Moles of ammonia = 0.469 M * 0.155 L = 0.0726 mol
Step 2: Calculate the concentration of hydroiodic acid added to the solution.
[HI] = 1.83 × 10^-2 mol / 0.155 L = 0.118 M
Step 3: Set up an ICE table to determine the equilibrium concentrations of all species.
| | Initial | Change | Equilibrium |
|:----------|:----------:|:----------:|:----------:|
| NH3 | 0.469 M | -x | (0.469 - x) M |
| HI | 0.118 M | -x | (0.118 - x) M |
| H3O+ | 0 | +x | x M |
| NH4+ | 0 | +x | x M |
Step 4: Write the equilibrium constant expression for the dissociation of hydroiodic acid.
Ka = [H3O+][I-] / [HI]
Step 5: From the ICE table and the Ka expression, set up an equation to solve for x.
Ka = x^2 / (0.118 - x)
10^-10 = x^2 / (0.118 - x)
Step 6: Solve for x.
x = 1.05 × 10^-6 M
Step 7: Calculate the pH of the solution.
pH = -log[H3O+]
= -log(1.05 × 10^-6)
= 5.98
Therefore, the pH of the solution after the addition of hydroiodic acid is 5.98.
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