A solution is 5.0× 10–5 M in each of these ions: Ag , SO42–, Cl–, CO32– Which precipitate will form?

Ag2SO4 (Ksp = 1.12× 10–5)
AgCl (Ksp = 1.77× 10–10)
Ag2CO3 (Ksp = 8.46× 10–12)

AgCl (Ksp = 1.77× 10–10)

Calculate Qsp for each salt and compare with Ksp.

For example, Qsp for AgCl = (5E-5)(5E-5) = 25E-10 = 2.5E-9.
Ksp = 1.77E-10 which is SMALLER than Qsp so AgCl will ppt. Remember the Ksp can't be exceeded. It the ion product IS exceeded, a ppt occurs.

Its AgCl can confirm

1.77

AgCl is the answer.

Definitely AgCl

To determine which precipitate will form, we need to compare the Q (ion product) values of each potential precipitate with their respective solubility product constant (Ksp) values.

The ion product (Q) is calculated by multiplying the concentrations of the individual ions involved in the reaction. In this case, we have three potential solubility reactions:

1) Ag2SO4 ↔ 2Ag+ + SO42–
2) AgCl ↔ Ag+ + Cl–
3) Ag2CO3 ↔ 2Ag+ + CO32–

Let's calculate the Q values for each of these reactions based on the given concentrations:

1) Q for Ag2SO4 = [Ag+]² × [SO42–]
Given: [Ag+] = [SO42–] = 5.0×10–5 M
Substituting the values, we get:
Q = (5.0×10–5)² × (5.0×10–5) = 2.5×10–13

2) Q for AgCl = [Ag+] × [Cl–]
Given: [Ag+] = [Cl–] = 5.0×10–5 M
Substituting the values, we get:
Q = (5.0×10–5) × (5.0×10–5) = 2.5×10–9

3) Q for Ag2CO3 = [Ag+]² × [CO32–]
Given: [Ag+] = [CO32–] = 5.0×10–5 M
Substituting the values, we get:
Q = (5.0×10–5)² × (5.0×10–5) = 1.25×10–14

Now let's compare the Q values with their respective Ksp values:

1) For Ag2SO4: Q = 2.5×10–13, Ksp = 1.12×10–5
Since Q < Ksp, no precipitate will form.

2) For AgCl: Q = 2.5×10–9, Ksp = 1.77×10–10
Since Q > Ksp, a precipitate of AgCl will form.

3) For Ag2CO3: Q = 1.25×10–14, Ksp = 8.46×10–12
Since Q < Ksp, no precipitate will form.

Based on the calculations, the precipitate that will form is AgCl.

qwerty

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