2. Does a precipitate of silver chloride form when 200.0 mL of 1.0 x 10-4 M AgNO3 (aq) and 900.0 mL of 1.0 x 10-6 M KCl (aq) are mixed at 25oC?[ Ksp AgCl @ 25oC = 1.6 x 10-10]
Calculate the pH of a NaOH solution of concentration 0.5M (Kw = 1.0 x 10-14 mol2dm-6 at 25oC)
You want to compare Ksp with Qsp.
(AgNO3) = 1E-4 x (200/1100) = ?M
(KCl) = 1E-6 x (900/1100) = ?M
Qsp = (Ag^+)(Cl^-) = ?
If Qsp > Ksp a ppt forms.
If Qsp < Ks no ppt forms.
5. Consider the following reaction: CO2 (g) + H2 (g) CO (g) + H2O (g)
Calculate the value of the equilibrium constant, Kc, for the above system, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapour were present in a 2.00 L reaction vessel at equilibrium.
You should post each question separately. Some won't answer posts that too long; i.e., consists of several questions.
Calculate the pH of a NaOH solution of concentration 0.5M (Kw = 1.0 x 10-14 mol2dm-6 at 25oC)
(H^+)(OH^-) = Kw.
Substitute and solve for (H^+) [note that (OH^-) = (NaOH)], then pH = -log(H^+)
5. Consider the following reaction: CO2 (g) + H2 (g) CO (g) + H2O (g)
Calculate the value of the equilibrium constant, Kc, for the above system, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapour were present in a 2.00 L reaction vessel at equilibrium.
I don't understand your problem with this. Substitute those numbers into the Kc. You want to substitute molarities; i.e., = mols/L. The moles are given; the volume is 2 L.
To determine if a precipitate of silver chloride (AgCl) will form when the solutions of silver nitrate (AgNO3) and potassium chloride (KCl) are mixed, we need to compare the ion concentrations to the solubility product constant (Ksp) for AgCl at 25oC.
Here are the steps to calculate if a precipitate will form:
1. Write the balanced chemical equation for the reaction between AgNO3 and KCl:
AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)
2. Determine the number of moles of AgNO3 and KCl in the solution:
- Moles of AgNO3 = volume (in L) × concentration (in M)
Moles of AgNO3 = 0.2000 L × 1.0 × 10^-4 M = 2.0 × 10^-5 mol
- Moles of KCl = volume (in L) × concentration (in M)
Moles of KCl = 0.9000 L × 1.0 × 10^-6 M = 9.0 × 10^-7 mol
3. Determine the common ion concentration:
The common ion in this case is chloride ions (Cl-), contributed by both AgCl and KCl.
- Concentration of Cl- = [KCl] = 1.0 × 10^-6 M
4. Calculate the ion product (Q), which is the product of the ion concentrations:
- Q = [Ag+] × [Cl-]
Q = (2.0 × 10^-5 mol / 1.1000 L) × (1.0 × 10^-6 mol / 1.1000 L) = 1.818 × 10^-10
5. Compare the ion product (Q) to the solubility product constant (Ksp):
If Q > Ksp, a precipitate will form. If Q < Ksp, no precipitate will form.
- Ksp for AgCl at 25oC = 1.6 × 10^-10
In this case, Q (1.818 × 10^-10) is greater than Ksp (1.6 × 10^-10), which means the ion product exceeds the solubility product. Therefore, a precipitate of silver chloride (AgCl) will form when the solutions of AgNO3 and KCl are mixed at 25oC.