Does any solid pbcl2 when 3.5mg nacl is dissolved in 0.25l of 0.12m pb(no3)2?
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To answer this question, we need to determine if a precipitate will form when 3.5 mg of NaCl is dissolved in 0.25 L of a 0.12 M Pb(NO3)2 solution.
To do this, we need to consider the solubility of PbCl2 and the concept of solubility product constant (Ksp).
1. Determine the balanced chemical equation:
Pb(NO3)2 (aq) + 2NaCl (aq) → PbCl2 (s) + 2NaNO3 (aq)
2. Write the expression for the solubility product constant (Ksp):
Ksp = [Pb2+][Cl-]^2
3. Look up the value of Ksp for PbCl2:
The Ksp value for PbCl2 is approximately 1.6 x 10^-5 at 25°C.
4. Calculate the concentration of Pb2+ ions:
Given that the initial concentration of Pb(NO3)2 is 0.12 M and the volume is 0.25 L, we can use the formula:
[Pb2+] = (moles of Pb(NO3)2) / (volume of solution in liters)
[Pb2+] = (0.12 mol/L * 0.25 L) / 1
[Pb2+] = 0.03 mol/L
5. Calculate the concentration of Cl- ions:
Given that we have 3.5 mg of NaCl, we need to convert it to moles:
moles of NaCl = (mass of NaCl in grams) / (molar mass of NaCl)
moles of NaCl = (3.5 mg / 1000) / 58.44 g/mol
moles of NaCl ≈ 6.0 x 10^-5 mol
Since NaCl dissociates into Na+ and Cl- ions in a 1:1 ratio, the concentration of Cl- ions is also 6.0 x 10^-5 mol/L.
6. Calculate the value of the solubility product (Qsp):
Qsp = [Pb2+][Cl-]^2
Qsp = (0.03 mol/L)(6.0 x 10^-5 mol/L)^2
Qsp ≈ 1.08 x 10^-9
7. Compare Qsp with Ksp:
If Qsp is greater than Ksp, then a precipitate is likely to form. However, if Qsp is less than Ksp, the solution remains unsaturated, and no precipitate will form.
In this case, Qsp ≈ 1.08 x 10^-9, which is significantly less than Ksp (1.6 x 10^-5). Therefore, no solid PbCl2 will form, as the solution remains unsaturated.