Suppose hydrogen sulfide is added to a solution that is 0.10 m in cu2+, pb2+, and ni2+ such that the concentration of h2s is 0.10 m. when the ph of the solution is adjusted to 1.00, a precipitate forms. what is the composition of the precipitate? h2s(aq) + 2h2o(l) 2h3o+(aq) + s2–(aq); kc = 1.1 × 10–20 salt ksp cus 6.0 × 10–36 pbs 2.5 × 10–27 nis 3.0 × 10–19
a. nis only
b. cus, pbs, and nis
c. cus and pbs
d. cus only
e. pbs and nis

C. CuS and PbS  

Explanation:

1. Calculate [H₃O⁺]

pH = 1.00  

2. Calculate the concentration of S²⁻

H₂S + 2H₂O ⇌ S²⁻ + 2H₃O⁺; Kc =  1.1 × 10⁻²⁰

3. Calculate Qsp for the sulfides

(a) CuS

CuS ⇌ Cu²⁺ + S²⁻; Ksp = 6.0 × 10⁻³⁶

Qsp = [Cu²⁺][S²⁻] = (0.10)(1.0 × 10⁻¹⁹) = 1.0 × 10⁻²⁰

Qsp > Ksp, so a precipitate of CuS will form.

(b) PbS

PbS ⇌ Pb²⁺ + S²⁻; Ksp = 2.5 × 10⁻²⁷

Qsp = [Pb²⁺][S²⁻] = (0.10)(1.0 × 10⁻¹⁹) = 1.0 × 10⁻²⁰

Qsp > Ksp, so a precipitate of PbS will form.

(iii) NiS

NiS ⇌ Ni²⁺ + S²⁻; Ksp = 3.0 × 10⁻¹⁹

Qsp = [Ni²⁺][S²⁻] = (0.10)(1.0 × 10⁻¹⁹) = 1.0 × 10⁻²⁰

Qsp < Ksp, so a precipitate will not form.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the answer is option a.