Question

Consider a tank initially containing 3L of 1M HCl. To this tank, we add 3L of 2M Ca(OH)₂, and allow the acid-base reaction to proceed to completion.

(a) At the end of this step, what are the concentrations of HCl, Ca(OH)₂ and CaCl₂ in the mixture ? (Please note that there is water generated during the reaction)

We then remove 2L of the contents of the tank, add 2L of 0.5M HCl, and allow the reaction (if any) to proceed to completion

(b) what are the concentrations of HCl, Ca(OH)₂ and CaCl₂ in the mixture now ? (Again, note that water may be generated due to the reaction)

[NB: You may assume that all of the above occurs at a temperature where the density of water is 1000 kg/m³ (1000 g/L)]

Please show work! I've missed class due to illness so I need to know how to do this.

Answer 1

(a) 2HCl + Ca(OH)₂ = CaCl₂ + 2H₂O

After Addition of 3L of 2M Ca(OH)₂, total volume = 6 L

After addition, concentration of HCl in the tank before reaction = 3 L * 1M/6 L = 0.5 M

Amount of HCl = 0.5 moles.L^­1 * 6000 mL /1000 mL = 3 moles

After addition, concentration of Ca(OH)₂ in the tank before reaction = 3 L * 2M/6 L = 1 M

Amount of Ca(OH)₂ = 1 moles.L^­1 *6000 mL/1000 mL = 6 moles

According to the reaction, 2 moles of HCl reacts with 1 moles of Ca(OH)₂. So 3 moles of HCl will react with 1.5 moles of Ca(OH)₂ and 1.5 moles of CaCl₂, 3 moles of water will be produced.

After the reaction,

Amount of HCl remains = 0 moles

Amount of Ca(OH)₂ remains = 6 - 1.5 = 4.5 moles

Amount of water produced = 3 moles = 3*18 g = 54 g = 0.054 kg/(1000 kg/m³)= 0.054 kg/(1 kg/dm³) =0.054 kg/(1 kg/L) = 0.054 L

Total volume of solution = 6+0.054 = 6.054 L

Concentration of HCl after reaction = 0

Concentration of Ca(OH)₂ after reaction = 4.5 moles * 1 L/6.054 L = 0.743 M

Concentration of CaCl₂ after reaction = 1.5 moles * 1L/6.054 L = 0.248 M

(b) 2 L solution of the tank contains,

HCl = 0 moles

Ca(OH)₂ = 0.743 moles.L^-1 * 2 L = 1.486 moles

CaCl₂ = 0.248 moles.L^-1 * 2 L = 0.496 moles

Amount of CaCl₂ remained in the tank = 1.5 - 0.496 = 1.004 moles

Amount of Ca(OH)₂ remained in the tank = 4.5 - 1.486 = 3.014 moles

Amount of HCl added = 0.5 moles.L^-1 * 2 L = 1 mole

Total volume of the solution = 6.054 L

According to the reaction, 2 moles of HCl reacts with 1 moles of Ca(OH)₂. So 1 mole of HCl will react with 0.5 moles of Ca(OH)₂ and 0.5 moles of CaCl₂, 1 mole of water will be produced.

After the reaction,

Amount of HCl remains = 0 moles

Amount of Ca(OH)₂ remains = 3.014-0.5 = 2.514 moles

Amount of CaCl₂ = 0.5 + 1.004 = 1.504 moles

Amount of water produced = 1 mole = 1*18 g = 18 g = 0.018 kg/(1000 kg/m³)= 0.018 kg/(1 kg/dm³) =0.018 kg/(1 kg/L) = 0.018 L

Total volume of solution = 6.054 L + 0.018 L = 6.072 L

Concentration of HCl after reaction = 0

Concentration of Ca(OH)₂ after reaction = 2.514 moles * 1 L/6.072 L = 0.414 M

Concentration of CaCl₂ after reaction = 1.504 moles * 1L/6.072 L = 0.248 M