A 20 mL of 0.25M HCN is mixed with 50 mL of 0.1 M aqueous NH3 . Calculate a. The degree of the resulting salt formed. b. the pH of the resulting solution. Given that Ka(HCN)=6.17×10^ 10 and Kb(NH3)=1.85×10^ 5

A 20 ml of 0.25m HCN is mixed with 50 ml of 0.1 m aqueous NH3 . Calculate a. The degree of the resulting salt formed. b. the ph of the resulting solution. Given that Ka(HCN)=6.17×10^-10 and Kb(NH3)=1.85×10^-5

Accepted Answer

Here's how to solve this problem:
Step 1: Determine the limiting reagent

HCN: 20 mL
0.25 mol/L = 5 mmol
NH3: 50 mL
0.1 mol/L = 5 mmol
Since both reactants are present in equal amounts, they will both be completely consumed in the reaction.
Step 2: Write the reaction and calculate the equilibrium constant
HCN(aq) + NH3(aq) ⇌ NH4+(aq) + CN-(aq)

K = Ka(HCN) / Kb(NH3) = (6.17 × 10^-10) / (1.85 × 10^-5) = 3.34 × 10^-5
Step 3: Calculate the initial concentrations of the products

[NH4+] = [CN-] = 5 mmol / (20 mL + 50 mL) = 0.0714 M
Step 4: Set up an ICE table to calculate the equilibrium concentrations
| | HCN | NH3 | NH4+ | CN- |
|-------------|-----------|-----------|------------|------------|
| Initial | 0 | 0 | 0.0714 | 0.0714 |
| Change | +x | +x | -x | -x |
| Equilibrium | x | x | 0.0714-x | 0.0714-x |
Step 5: Solve for x using the equilibrium constant
K = [NH4+][CN-] / [HCN][NH3] = (0.0714-x)^2 / x^2 = 3.34 × 10^-5
Solving for x, we get x = 0.0698 M
Step 6: Calculate the degree of salt formation

Degree of salt formation = (concentration of salt formed) / (initial concentration of limiting reagent) = (0.0714 - 0.0698) / 0.0714 = 0.0225
Step 7: Calculate the pH of the resulting solution
Since the reaction produces a buffer solution of NH4+ and CN-, we can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log([CN-] / [NH4+]) = -log(6.17 × 10^-10) + log(0.0698 / 0.0714) = 9.21
Therefore:

a. The degree of salt formation is 0.0225.
b. The pH of the resulting solution is 9.21.

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