HCl is slowly added to a solution that is .250 M in Pb^2+ and .00150 M in Ag+. Which precipitate forms first, PbCl2 or AGCl? At what Cl- ion concentration will the precipitate first appear? Ksp for PbCl2= 1.6*10^-5, Ksp for AgCl= 1.7*10^-10
Seriously don't understand Ksp and solubility. How do I even approach this problem?
PbCl2 ==> Pb^++ + 2Cl^-
AgCl --> Ag^+ + Cl^-
Using Ksp for PbCl2, calculate what the (Cl^-) must be to ppt PbCl2. You know (Pb^++) and Ksp, solve for (Cl^-).
Do the same for AgCl. You know Ag^+ and Ksp, solve for (Cl^-). The one with the lower Cl^- will appear first so that answers the first two questions. Do that and perhaps the third question will be easier than it is now.
To determine which precipitate forms first and at what concentration, we need to compare the solubility product constants (Ksp) for PbCl2 and AgCl. The precipitate will form when the product of the concentrations of the ions involved exceeds the value of Ksp for each compound.
To approach this problem, we can follow these steps:
Step 1: Write the balanced chemical equations for the dissociation of PbCl2 and AgCl in water.
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Step 2: Calculate the theoretical concentrations of the Pb2+ and Ag+ ions in the solution.
Given that the solution is 0.250 M in Pb^2+ and 0.00150 M in Ag+, these are the initial concentrations of the ions.
Step 3: Calculate the ion concentrations after adding HCl.
Since HCl dissociates completely in water, it will increase the concentration of Cl- ions in the solution. This change in concentration must be taken into account.
Step 4: Calculate the ion concentrations at which the precipitates will first appear.
This is done by comparing the product of ion concentrations ([Pb2+][Cl-] for PbCl2 and [Ag+][Cl-] for AgCl) to the respective Ksp values.
The compound with the smaller Ksp value will have a lower solubility and will form a precipitate first.
Step 5: Calculate the Cl- ion concentration at which the precipitate first appears.
At this concentration, the product of the ion concentrations must equal the respective Ksp value.
Let's go through the calculations step by step:
Step 1:
The balanced chemical equations for the dissociation of PbCl2 and AgCl in water are:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Step 2:
The initial concentration of Pb2+ is 0.250 M.
The initial concentration of Ag+ is 0.00150 M.
Step 3:
When HCl is added, it dissociates to provide Cl- ions. Let's assume x mol/L of HCl dissociates to form x mol/L of Cl- ions.
The concentration of Cl- ions will then be:
0.00150 M (Ag+) + x M (Cl-) for AgCl formation
x M (Cl-) for PbCl2 formation
Step 4:
Using the solubility product constant expressions for PbCl2 and AgCl, we can write:
[Pb2+][Cl-]^2 = Ksp for PbCl2
[Ag+][Cl-] = Ksp for AgCl
Comparing the product of ion concentrations to the respective Ksp values:
For AgCl:
(0.00150 + x)(x) = Ksp for AgCl
x^2 + 0.00150x = Ksp for AgCl ...(Equation 1)
For PbCl2:
(0.250)(x) = Ksp for PbCl2
0.250x = Ksp for PbCl2 ...(Equation 2)
Step 5:
To find the Cl- ion concentration at which the precipitate first appears, we need to solve both equations and find the value of x, which represents the concentration of Cl- ions at the precipitation point.
Solve Equation 1 for x:
x^2 + 0.00150x - Ksp for AgCl = 0
Solve Equation 2 for x:
0.250x - Ksp for PbCl2 = 0
By solving the above quadratic equations, we can find the values of x.
After finding the value of x, you can determine which precipitate forms first by comparing the solubility products of PbCl2 and AgCl. The compound with the smaller Ksp value will form a precipitate first.
I hope this step-by-step explanation helps! If you have any further questions, please feel free to ask.
To determine which precipitate, PbCl2 or AgCl, forms first and at what Cl- ion concentration the first precipitate appears, we need to consider the solubility product constant (Ksp) and solubility of each compound.
Ksp is an equilibrium constant that represents the solubility of a sparingly soluble salt in water. In general, when the concentration of the ions in a solution exceeds the Ksp of a particular compound, the compound will precipitate. The greater the Ksp value, the more soluble the compound.
In this problem, the Ksp values given are:
Ksp for PbCl2 = 1.6 × 10^-5
Ksp for AgCl = 1.7 × 10^-10
Before we proceed, we need to determine the molar solubility of each compound, which is the concentration of the ions in a saturated solution.
The molar solubility of PbCl2 (s) is given by:
PbCl2 (s) ⇌ Pb^2+ (aq) + 2 Cl- (aq)
Using the stoichiometry of the balanced equation, the molar solubility can be represented as [Pb^2+] = [Cl-] = s (assuming complete dissociation). Therefore, [Pb^2+] = s and [Cl-] = 2s.
The molar solubility of AgCl (s) is given by:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Using the stoichiometry of the balanced equation, the molar solubility can be represented as [Ag+] = [Cl-] = s (assuming complete dissociation). Therefore, [Ag+] = s and [Cl-] = s.
Now, we can set up the expressions for the Ksp of each compound:
Ksp for PbCl2 = [Pb^2+] [Cl-]^2
Ksp for AgCl = [Ag+] [Cl-]
Substituting the expressions for [Pb^2+] and [Cl-] for PbCl2:
Ksp for PbCl2 = s (2s)^2 = 4s^3
Substituting the expressions for [Ag+] and [Cl-] for AgCl:
Ksp for AgCl = s (s) = s^2
To determine which precipitate forms first, we need to compare the Ksp values. PbCl2 has a larger Ksp (1.6 × 10^-5) compared to AgCl (1.7 × 10^-10), which indicates that PbCl2 is more soluble than AgCl. Therefore, AgCl will precipitate first.
To find the Cl- ion concentration at which the first precipitate appears, we need to find the solubility of AgCl using its Ksp.
Ksp for AgCl = s^2
1.7 × 10^-10 = s^2
Taking the square root of both sides gives:
s = sqrt(1.7 × 10^-10)
Now, we can find the Cl- ion concentration:
Cl- concentration = 2s (since [Cl-] for both compounds is s)
Cl- concentration = 2 * sqrt(1.7 × 10^-10)
Evaluate the expression to find the Cl- ion concentration.