# 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?

Aug 2, 2017

$\text{molarity} = 0.331$ $\text{mol/L}$

#### Explanation:

We're asked to find the final concentration of the $\text{HCl}$ solution after two separate $\text{HCl}$ solutions are mixed.

To do this, we can use the molarity equation:

$\text{molarity" = "mol HCl"/"L solution}$

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

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

where

• ${\text{mol}}_{1}$ and ${\text{mol}}_{2}$ are the number of moles of $\text{HCl}$ in solutions $1$ and $2$

• ${\text{L}}_{1}$ and ${\text{L}}_{2}$ are the volumes, in liters, of the two $\text{HCl}$ solutions

We're actually going to use the molarity equation to find the number of moles of $\text{HCl}$ in each solution:

Solution 1:

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

Solution 2:

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

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

$\left[H C l\right] \approx 0.3 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We use the relationship, $\text{Concentration"="Moles of solute"/"Volume of solution}$.

We assume (reasonably) that the volumes are additive, and we work out the entire number of moles of $H C l$.........

$\text{Solution 1: Moles of HCl}$ $= 16.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.130 \cdot m o l \cdot {L}^{-} 1 = 2.08 \times {10}^{-} 3 \cdot m o l$

$\text{Solution 2: Moles of HCl}$ $= 12.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.600 \cdot m o l \cdot {L}^{-} 1 = 7.20 \times {10}^{-} 3 \cdot m o l$

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

$\frac{2.08 \times {10}^{-} 3 \cdot m o l + 7.20 \times {10}^{-} 3 \cdot m o l}{16.0 \times {10}^{-} 3 \cdot L + 12.0 \times {10}^{-} 3 \cdot L} = 0.331 \cdot m o l \cdot {L}^{-} 1$.