A #=16.0*mL# volume of #0.130*mol*L^-1# #HCl(aq)# was mixed with a #12.0*mL# volume of #0.600*mol*L^-1# #HCl(aq)#. What is the resultant concentration?

2 Answers
Aug 2, 2017

Answer:

#"molarity" = 0.331# #"mol/L"#

Explanation:

We're asked to find the final concentration of the #"HCl"# solution after two separate #"HCl"# solutions are mixed.

To do this, we can use the molarity equation:

#"molarity" = "mol HCl"/"L solution"#

Since we're combining two separate solutions, this can also be represented as

#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#

where

  • #"mol"_1# and #"mol"_2# are the number of moles of #"HCl"# in solutions #1# and #2#

  • #"L"_1# and #"L"_2# are the volumes, in liters, of the two #"HCl"# solutions

We're actually going to use the molarity equation to find the number of moles of #"HCl"# in each solution:

Solution 1:

#"mol"_1 = ("molarity"_1)("L"_1) = (0.130"mol"/(cancel("L")))overbrace((0.0160cancel("L")))^"converted to liters" = color(red)(ul(0.00208color(white)(l)"mol HCl"#

Solution 2:

#"mol"_2 = ("molarity"_2)("L"_2) = (0.600"mol"/(cancel("L")))overbrace((0.0120cancel("L")))^"converted to liters" = color(green)(ul(0.00720color(white)(l)"mol HCl"#

The molarity of the final solution is thus

#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#

#"molarity" = (color(red)(0.00208color(white)(l)"mol") + color(green)(0.00720color(white)(l)"mol"))/(0.0160color(white)(l)"L" + 0.0120color(white)(l)"L") = color(blue)(ulbar(|stackrel(" ")(" "0.331color(white)(l)"mol/L"" ")|)#

Aug 2, 2017

Answer:

#[HCl]~~0.3*mol*L^-1#

Explanation:

We use the relationship, #"Concentration"="Moles of solute"/"Volume of solution"#.

We assume (reasonably) that the volumes are additive, and we work out the entire number of moles of #HCl#.........

#"Solution 1: Moles of HCl"# #=16.0*mLxx10^-3*L*mL^-1xx0.130*mol*L^-1=2.08xx10^-3*mol#

#"Solution 2: Moles of HCl"# #=12.0*mLxx10^-3*L*mL^-1xx0.600*mol*L^-1=7.20xx10^-3*mol#

And thus the final concentration is given by the quotient......

#(2.08xx10^-3*mol+7.20xx10^-3*mol)/(16.0xx10^-3*L+12.0xx10^-3*L)=0.331*mol*L^-1#.