Consider a tank initially containing 3L of 1M HCl. To this tank, we add 3L of 2M

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)]

Show work please.

Explanation / Answer

Ca(OH)2 + 2HCl ----> CaCl2 + H2O

Moles of HCl = 3 x 1 = 3

Moles of Ca(OH)2 = 3 x 2 = 6

Moles of Ca(OH)2 reacted = 3 / 2 = 1.5 moles

(a) Conc of HCl = 0

[Ca(OH)2] = (6 - 1.5) / (3 + 3) = 0.75 M

[CaCl2] = 1.5 / 6 = 0.25 M

Remaining = 6 - 2 = 4 liters

Moles of HCl added = 2 x 0.5 = 1 mole

moles of Ca(OH)2 = 4 x 0.75 = 3 moles

moles of CaCl2 = 0.25 x 4 = 1

Moles of Ca(OH)2 reacted = 0.5

Moles of Ca(OH)2 remaining = 3 - 0.5 = 2.5 moles

Moles of CaCL2 formed = 0.5

Total moles of CaCl2 = 0.5 + 1 = 1.5 moles

Moles of HCl left = 0

[Ca(OH)2] = 2.5 / (4 + 2) = 0.417 M

[CaCl2] = 1.5 / 6 = 0.25 M

[HCl] = 0

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