Please judge my answer
Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)
a) Calculate the ion product of the potential precipitate.
b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 10-10.
Answer:
For a, doing the ion product is nessecary to get the answer. For part b of this question, comparing the ion product with Ksp is nessecary. Determining the ion product consists of using the formula Kw=Ka/Kb, for Ka I would do 0.0045/0.0085 = 0.529 and for Kb, I would do 0.45/0.0135 = 33.33. I would then use the formula Kw = Ka/Kb to get a value of 1.0 x 10^-2. This is my ion product value. When comparing the ion product to Ksp, I see that 1.0 x 10^-2 is greater than 1.8 x 10^-10, therefore a precipitate will form.
You are correct that a ppt of AgCl will form BUT you didn't arrive at the answer correctly. If I were grading your paper I would give you no credit.
There is no Ka or Kb to worry about. Not with Ksp. Ksp is what we are concerned with.
mol AgNO3 = M x L = 0.45 M 0.045 L = ??
mol NaCl = M x L = 0.0135 M x 0.085 L = ??
(Ag^+) = mols/total volume
total volume = 85 mL + 45 mL = 130 mL = 0.130 L.
(Cl^-) = mols/total volume.
ion produce is (Ag^+) from above *(Cl^-) from above. Then compare with Ksp.
It seems like your answer contains some incorrect information and calculations. Let me guide you on how to correctly calculate the ion product and determine whether a precipitate will form.
a) To calculate the ion product of the potential precipitate, you need to find the concentrations of the ions involved in the reaction, which are silver ions (Ag+) and chloride ions (Cl-).
Given:
Volume of AgNO3 solution = 45 mL
Molarity of AgNO3 solution = 0.45 M
Volume of NaCl solution = 85 mL
Molarity of NaCl solution = 1.35 x 10^-2 M
To find the concentration of Ag+ ions:
Concentration of Ag+ = Volume of AgNO3 solution x Molarity of AgNO3 solution
= 45 mL x 0.45 M
= 20.25 mmol (millimoles)
To find the concentration of Cl- ions:
Concentration of Cl- = Volume of NaCl solution x Molarity of NaCl solution
= 85 mL x 1.35 x 10^-2 M
= 1.1475 mmol
Now, we can calculate the ion product (Q) by multiplying the concentrations of the ions:
Q = [Ag+] x [Cl-]
= 20.25 mmol x 1.1475 mmol
= 23.239 mmol^2
b) To determine whether a precipitate will form, we compare the ion product (Q) with the solubility product constant (Ksp) of AgCl.
Given:
Ksp of AgCl = 1.8 x 10^-10
If the ion product (Q) is greater than the solubility product constant (Ksp), a precipitate will form. Conversely, if Q is less than Ksp, no precipitate will form.
In our case, Q = 23.239 mmol^2, and Ksp = 1.8 x 10^-10. Since Q > Ksp, it means the ion product is greater than the solubility product constant, and a precipitate of AgCl will form.
Therefore, the correct answer would be:
a) The ion product of the potential precipitate is 23.239 mmol^2.
b) Yes, a precipitate will form because the ion product (Q) is greater than the solubility product constant (Ksp).