In quality control of raw materials for pharmaceutical production often used foul reactions to prove that raw materials in salt form containing the appropriate salt. Hydrochloride (eg. Amitriptyline hydrochloride) can be detected for example by precipitating the chloride ions like chloride. Chloride is sparingly soluble salt having solubility product, Ksp (AgCl) 1.6 · 10-10 at 25 ° C.
a) Calculate the molar solubility of the silver chloride at 25 ° C.
b) there will be a precipitation of silver chloride in 10 mL of a solution of 20 mM amitriptyline hydrochloride is added 1 mL of 2 M silver nitrate (AgNO3). Justify your answer with calculations.
a) To calculate the molar solubility of silver chloride (AgCl) at 25°C, we can use the solubility product constant (Ksp) for AgCl.
The general equation for the dissolution of AgCl is:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
The Ksp expression for this reaction is:
Ksp = [Ag+] [Cl-]
At equilibrium, the concentrations of Ag+ and Cl- are equal, since AgCl has a 1:1 stoichiometry. Therefore, we can represent their concentration as x, giving us:
Ksp = x * x = x^2
The given value for Ksp of AgCl is 1.6 × 10^-10 at 25°C.
Setting up the equation:
1.6 × 10^-10 = x^2
Taking the square root of both sides:
x = √(1.6 × 10^-10)
Using a calculator:
x ≈ 1.26 × 10^-5
Therefore, the molar solubility of silver chloride at 25°C is approximately 1.26 × 10^-5 M.
b) To determine if precipitation of silver chloride will occur when 1 mL of 2 M silver nitrate (AgNO3) is added to 10 mL of a 20 mM amitriptyline hydrochloride solution, we need to compare the ionic product (Q) to the solubility product constant (Ksp) of AgCl.
First, let's determine the initial concentration of chloride ions (Cl-) in the solution of amitriptyline hydrochloride:
20 mM = 20 × 10^-3 M
Since the amitriptyline hydrochloride dissociates into one chloride ion (Cl-) for every molecule, the concentration of chloride ions is also 20 × 10^-3 M.
Next, let's calculate the concentration of silver ions (Ag+) that will be present after adding 1 mL of 2 M silver nitrate (AgNO3):
Volume of AgNO3 added = 1 mL = 1 × 10^-3 L
Concentration of Ag+ from AgNO3 = 2 M
Using the dilution equation:
Concentration of Ag+ in 1 mL = (2 M) × (1 × 10^-3 L / 11 × 10^-3 L)
Simplifying:
Concentration of Ag+ in 1 mL = 0.18 M
Now we can calculate the ionic product (Q) using the concentrations of Ag+ and Cl- from the previous calculations:
Q = [Ag+] [Cl-] = (0.18 M) × (20 × 10^-3 M) = 3.6 × 10^-3
Comparing the ionic product (Q) to the solubility product constant (Ksp):
Q = 3.6 × 10^-3
Ksp = 1.6 × 10^-10
Since Q > Ksp, the ionic product exceeds the solubility product, indicating that silver chloride will precipitate.
Therefore, precipitation of silver chloride will occur when 1 mL of 2 M silver nitrate (AgNO3) is added to 10 mL of a 20 mM amitriptyline hydrochloride solution.
To solve this problem, we need to calculate the molar solubility of silver chloride (AgCl) at 25 °C and determine whether precipitation of silver chloride will occur when 1 mL of 2 M silver nitrate (AgNO3) is added to a 10 mL solution of 20 mM amitriptyline hydrochloride.
Let's break down the steps to find the molar solubility of silver chloride and perform the necessary calculations:
a) Calculate the molar solubility of silver chloride at 25 °C:
Step 1: Write the solubility product expression for silver chloride (AgCl):
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The solubility product expression can be represented as:
Ksp = [Ag+][Cl-]
Step 2: Use the solubility product constant (Ksp) to calculate the molar solubility of silver chloride.
Ksp = [Ag+][Cl-] = (x)(x) = x^2
where x represents the molar solubility of AgCl.
Ksp value provided: Ksp (AgCl) = 1.6 × 10^-10
Let's solve for x:
1.6 × 10^-10 = x^2
Taking the square root of both sides:
x = sqrt(1.6 × 10^-10)
x ≈ 1.26 × 10^-5
Therefore, the molar solubility of silver chloride at 25 °C is approximately 1.26 × 10^-5 M.
b) Determine if precipitation of silver chloride will occur when 1 mL of 2 M silver nitrate (AgNO3) is added to a 10 mL solution of 20 mM amitriptyline hydrochloride:
Step 1: Calculate the moles of silver nitrate added:
Moles of AgNO3 = volume (L) × molarity (mol/L)
Moles of AgNO3 = 0.001 L × 2 mol/L
Moles of AgNO3 = 0.002 mol
Step 2: Calculate the initial concentration of silver ions (Ag+) in the 1 mL solution of 2 M AgNO3:
Initial concentration of Ag+ ions = Moles of AgNO3 / Volume of solution (L)
Initial concentration of Ag+ ions = 0.002 mol / 0.001 L
Initial concentration of Ag+ ions = 2 M
Step 3: Calculate the concentration of chloride ions (Cl-) in the 10 mL solution of 20 mM amitriptyline hydrochloride:
Concentration of Cl- ions = 20 mM = 20 × 10^-3 M
Step 4: Compare the initial concentration of Ag+ ions with the concentration of Cl- ions:
If [Ag+] < [Cl-], precipitation of AgCl will occur. Otherwise, no precipitation will occur.
[Ag+] = 2 M
[Cl-] = 20 × 10^-3 M
Since [Ag+] is greater than [Cl-], precipitation of AgCl will occur when 1 mL of 2 M AgNO3 is added to the 10 mL solution of 20 mM amitriptyline hydrochloride.
In conclusion, the molar solubility of silver chloride at 25 °C is approximately 1.26 × 10^-5 M, and precipitation of silver chloride will occur when 1 mL of 2 M silver nitrate (AgNO3) is added to a 10 mL solution of 20 mM amitriptyline hydrochloride.