determine the concentration of NH3(aq) that is required to dissolve 775 mg of AgCl(s) in 100.0 mL of solution. The Ksp of AgCl is 1.77× 10–10.]
please help!!!!!!!
To determine the concentration of NH3(aq) required, we need to use the concept of solubility product constant (Ksp).
Let's break down the steps to solve this problem:
Step 1: Write the balanced equation for the dissociation of silver chloride (AgCl).
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
Step 2: Write the expression for the solubility product constant (Ksp) of AgCl.
Ksp = [Ag+][Cl–]
Step 3: Use stoichiometry to determine the concentration of Cl– in the solution.
Since AgCl dissociates into one Ag+ ion and one Cl– ion:
[Cl–] = [Ag+] = x (let's assume the concentration of Cl– is x)
Step 4: Substitute the values into the Ksp expression and solve for x.
Ksp = x^2
1.77×10–10 = x^2
Taking the square root of both sides, we get:
x = √(1.77×10–10)
x = 1.33×10–5 M
Step 5: Since each AgCl dissociates into one Cl– ion, the concentration of NH3(aq) should be the same to keep the solution electrically neutral.
Therefore, the concentration of NH3(aq) required to dissolve 775 mg of AgCl(s) in 100.0 mL can be calculated as follows:
[(mass of AgCl)/(molar mass of NH3)] / (volume of solution in liters)
[(775 mg)/(143.32 g/mol)] / (0.100 L) = 54.0 mM
So, the concentration of NH3(aq) required to dissolve 775 mg of AgCl(s) in 100.0 mL of solution is 54.0 mM.
To determine the concentration of NH3(aq) required to dissolve AgCl(s), we can use the concept of the common ion effect. When NH3(aq) reacts with AgCl(s), it forms Ag+(aq) and Cl-(aq) ions in solution.
First, let's write the balanced equation for the dissolution of AgCl(s) in NH3(aq):
AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq)
Now, let's set up an equilibrium expression using the equilibrium constant (Ksp) for AgCl:
Ksp = [Ag+(aq)][Cl-(aq)]
Since Ag(NH3)2+(aq) is a complex ion, its concentration is related to the concentration of NH3(aq) through the formation constant (Kf):
[Ag(NH3)2+(aq)] = Kf * [Ag+(aq)] * [NH3(aq)]^2
Now, we can substitute the expressions for [Ag+(aq)] and [Cl-(aq)] into the Ksp expression in terms of NH3(aq) concentration:
Ksp = ([Ag(NH3)2+(aq)] / [NH3(aq)]^2) * [NH3(aq)]^2
Given that Ksp = 1.77×10^(-10), we can substitute the values and solve for [NH3(aq)]^2:
1.77×10^(-10) = (Kf * [Ag+(aq)] * [NH3(aq)]^2) / [NH3(aq)]^2
Simplifying the equation, we get:
1.77×10^(-10) = Kf * [Ag+(aq)]
Now, let's convert the mass of AgCl to moles to find the concentration of Ag+:
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = 775 mg = 0.775 g
Number of moles of AgCl = (0.775 g) / (107.87 g/mol)
Since 1 mol of AgCl dissociates to form 1 mol of Ag+, the concentration of Ag+ is equal to [Ag+]:
[Ag+] = (0.775 g / 107.87 g/mol) / (0.100 L)
Now, substitute the value of [Ag+] in the equation:
1.77×10^(-10) = Kf * [(0.775 g / 107.87 g/mol) / (0.100 L)]
Finally, solve for the concentration of NH3(aq):
[NH3(aq)] = √(1.77×10^(-10) / Kf * [(0.775 g / 107.87 g/mol) / (0.100 L)])
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