Question

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?

Answer 1

`"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"" ")|)`

Answer 2

`[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`.