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
Explanation:
We're asked to find the final concentration of the
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
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"" "))#
Explanation:
We use the relationship,
We assume (reasonably) that the volumes are additive, and we work out the entire number of moles of
And thus the final concentration is given by the quotient......