# How many "mL" of 2.15% (by mass) "HCl" would be required to neutralize "375 mL" of "165 mEq/L" "Ba(OH)"_2?

## Assume all solutions have a density of $1.12 \text{g"/"mL}$. $2 \text{HCl (aq)"+"Ba(OH)"_2"(aq)"\to2"H"_2"O (l)"+"BaCl"_2"(aq)}$

Jul 17, 2018

This has been answered before with slightly different numbers. It may be time for you to go back and refresh your memory on what we talked about there for practice.

I get that $\text{93.8 mL}$ of $\text{0.660 M HCl}$ reacts with $\text{375 mL}$ of "0.0825 M Ba"("OH")_2.

As before, a milliequivalent is just the number of millimols with respect to ${\text{OH}}^{-}$ for a strong base or ${\text{H}}^{+}$ for a strong acid. All we have different here are the numbers.

What we do is convert the $\text{mEq/L}$ to $\text{mol/L}$ of ${\text{OH}}^{-}$, and convert the $\text{%w/w}$ of $\text{HCl}$ to $\text{mol/L}$ as well.

BASE CONCENTRATION

As mentioned, ${\text{1 Eq" = "1 mol OH}}^{-}$ in strong bases or ${\text{H}}^{+}$ in strong acids. So:

["OH"^(-)] = (165 cancel"m"cancel"Eq" cancel("Ba"("OH")_2))/"L" xx (1 cancel"thing")/(1000 cancel"milli"cancel"things") xx ("2 mol OH"^(-))/(2 cancel"Eq" cancel("Ba"("OH")_2))

$= \underline{{\text{0.165 M OH}}^{-}}$

(NOTE: since there are approximately two ${\text{OH}}^{-}$ for every one "Ba"("OH")_2, assuming 100% dissociation, ["OH"^(-)] ~~ 2["Ba"("OH")_2], so ["Ba"("OH")_2] ~~ "0.0825 M".)

Here we simply treated "milli" as a separate object.

For instance, $\text{1 mmol"/"mL}$ is the same as $\text{1 milli"/"1 milli" xx "1 mol"/"L}$.

ACID CONCENTRATION

Percent by mass, or $\text{%w/w}$, is the mass of solute in $\text{100 g}$ of the system:

2.15%"w/w HCl" = "2.15 g HCl"/"100 g solution"

So, use the molar mass of $\text{HCl}$ to get to $\text{mols}$, and use the density of the solution to get to $\text{L}$, thereby giving you $\text{mol/L}$. Since $\text{HCl}$ is a strong acid:

$\left[\text{H"^(+)] = ["HCl}\right]$

and so, treating the numerator and denominator separately:

$\left[\text{H"^+] = (2.15 cancel"g HCl" xx ("1 mol")/(36.461 cancel"g HCl"))/(100 cancel"g solution" xx cancel"mL"/(1.12 cancel"g") xx "L"/(1000 cancel"mL}\right)$

$= \underline{{\text{0.660 M H}}^{+}}$

To evaluate this, just evaluate the top and bottom separately and then you should get $\text{0.0590 mols}$ of $\text{HCl}$ dissolved in $\text{0.0893 L}$.

ACID/BASE REACTION

Normally we would need a mol ratio from the chemical reaction, but since everything is now in terms of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$, we just react them exactly.

We know the ${\text{OH}}^{-}$ is contained in $\text{375 mL}$, so the mols are:

$0.375 {\cancel{\text{L" xx "0.165 mol"/cancel"L" = "0.0619 mols OH}}}^{-}$

And we know we want exact reaction, so we want ${\text{0.0619 mols H}}^{+}$. That is contained in:

color(blue)(V_(HCl)) = "L"/(0.660 cancel("mol H"^(+))) xx 0.0619 cancel("mols H"^(+))

= "0.0938 L" = color(blue)("93.8 mL")