Predict whether a precipitate will form from each of the following combinations: (a) 25 ml of 1 × 10−5 m co(no3)2 and 75 ml of 5 × 10−4 m na2s; (b) 500 ml of 7.5 × 10−4 m alcl3 and 100 ml of 1.7 × 10−5 m hg2(no3)2; (c) 1.5 l of 0.025 m bacl2 and 1.25 l of 0.014 m pb(no3)2

(a) Yes; (b) yes; (c) no  

Step-by-step explanation:

(a) CoS

The solutions dilute each other on mixing.

Total volume is 100 mL.

[Co²⁺] =  1 × 10⁻⁵ × 25/100 = 2.5 × 10⁻⁶ mol·L⁻¹

 [S²⁻] = 5 × 10⁻⁴ × 75/100  = 3.8 × 10⁻⁴ mol·L⁻¹

               CoS ⇌       Co²⁺     +     S²⁻; K_sp = 4 × 10⁻²¹

I/mol·L⁻¹:               2.5 × 10⁻⁶    3.8 × 10⁻⁴

Q_sp = [Co²⁺][S²⁻] = 2.5 × 10⁻⁶ × 3.8 × 10⁻⁴ = 9.4 × 10⁻¹⁰

Q_sp > K_sp. ∴ A precipitate will form.

(b) Hg₂Cl₂

Total volume is 600 mL.

    [Cl⁻] = (3 × 7.5× 10⁻⁴) × 500/600 = 1.88 × 10⁻³ mol·L⁻¹

[Hg₂²⁺] =        1.7 × 10⁻⁵  ×  100/600 = 2.83 × 10⁻⁶ mol·L⁻¹

               Hg₂Cl₂ ⇌       Hg₂²⁺     +    2Cl⁻; K_sp = 1.3 × 10⁻¹⁸

I/mol·L⁻¹:                  2.83 × 10⁻⁶    1.88 × 10⁻³

Q_sp = [Hg₂²⁺][Cl⁻]²   = 2.83 × 10⁻⁶ × (1.88 × 10⁻³)² = 1.0 × 10⁻¹¹

Q_sp > K_sp. ∴ A precipitate will form.

(c) PbCl₂

Total volume is 2.75 L.

 [Cl⁻] = (2 × 0.025) ×   1.5/2.75 = 2.73 × 10⁻² mol·L⁻¹

[Pb²⁺] =     0.014      × 1.25/2.75 = 6.36 × 10⁻³ mol·L⁻¹

                PbCl₂ ⇌      Pb²⁺     +      2Cl⁻; K_sp = 1.6 × 10⁻⁵

I/mol·L⁻¹:                6.36 × 10⁻³   2.73 × 10⁻²

Q_sp = [Pb²⁺][Cl⁻]²   = 6.36 × 10⁻³ × (2.73 × 10⁻²)² = 4.7 × 10⁻⁶

Q_sp < K_sp. ∴ A precipitate will not form.

for points guys

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 :