A 25.0 mL aliquot of concentated hydrochloric acid (11.7 M) is added to 175.00 m

Question

A 25.0 mL aliquot of concentated hydrochloric acid (11.7 M) is added to 175.00 mL of 3.25 M hydrochloric acid. (a) Is this a dilution? (b)Determine the number of moles of hydrochloric acid from the 175.00 mL of 3.25 M hydrochloric acid. (c) Determine the number of moles of hydrochlocic acid in the solution after the addition of the concentrated hydrochloric acid. Asume the volumes are additive. (d) Detemine the molarity of the solution after the addition of the concentrated hydrochloric acid. assume the volumes are additive

Explanation / Answer

a)It is dilution from 11.7M HCl point of view while 3.25M HCl is getting concentrated.

b)Moles of hydrochloric acid in 175ml of 3.25 =3.25*175/1000 =0.57 moles

Moles of hydrochloric acid in 11.7ml of 25ml =11.7*25/1000 =0.2925 moles

c)Total moles of Hydrochloric acid = 0.57+0.2925= 0.8625 moles

Volume of mixed solution = 25+175= 200ml =0.2L

d) Molarity = moles of hydrochloric acid/ volume in L= 0.8625/0.2 =4.3125 (so it is dilution of 11.7 M HCl and concentration of 3.25M HCl)

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