Given concentrated hydrochloric at #10.6*mol*L^-1#, how do we prepare a #2*L# volume of #HCl# at #2.5*mol*L^-1# concentration?

2 Answers
anor277
Jan 15, 2018

We use conc. hydrochloric acid....#10.6*mol*L^-1#...and remember YOU ALWAYS ADD ACID TO WATER.....

Explanation:

We want #2*Lxx2.5*mol*L^-1=5.0*mol# with respect to #HCl#.

And so we take ..............#(5.0*mol)/(10.6*mol*L^-1)xx1000*mL*L^-1=472*mL#

...and this solution is added to #1526*mL# distilled water.....

The order of addition is IMPORTANT.....

#"If you spit in acid, it SPITS BaCK!!"#

Truong-Son N.
Jan 15, 2018

If you have #"2 L"# of a #"2.5 mol/L"# solution of #"HCl"# in water, then you simply need #"5.0 mols"# of #"HCl"# within twice of #"1 L"# volume:

#"2.5 mol"/cancel"L" xx 2 cancel"L" = "5.0 mols HCl"#

Therefore, the mass needed is:

#5.0 cancel"mols HCl" xx "36.4609 g HCl"/cancel"1 mol" = color(red)"182.3 g HCl"#

Hypothetically, if for some magical reason one had obtained #"HCl"(s)# at #"298.15 K"# and #"1 atm"#, then one would attempt to dissolve this much mass in some water, and then fill up to the #"2-L"# mark.

This is however, #color(red)"impractical"#. Actual reagent-grade #"HCl"# is usually about #37% "w/w"#, or about #"12.06 M"#. So you would have to dilute it down from that.

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