# How many "mL" of 1.25% (by mass) "HCl" would be required to neutralize "215 mL" of "325 mEq/L" "Ba"("OH")_2?

## Assume all solutions have a density of 1.08 g/mL. Note: My issue is understanding how to apply equivalents (unit Eq) to acids and bases. With singular elements/molecules, I understand, but not compounds.

##### 2 Answers
Jul 4, 2018

The volume needed is $\text{189 mL}$ of 12.5%"w/w HCl".

Since we used equivalents, we don't need to worry about mol ratios here, but had we used mols instead, we would.

Equivalents are defined with respect to the ${\text{OH}}^{-}$ or ${\text{H}}^{+}$ in a strong base or strong acid, since they both have charge magnitudes of $1$. "Ba"("OH")_2 is considered a strong base (its solubility at ${25}^{\circ} \text{C}$ is $\text{0.1077 M}$).

Suppose you have "0.1000 M Ba"("OH")_2. In that case, you would have "0.1000 mol/L Ba"("OH")_2, or "0.2000 Eq/L Ba"("OH")_2, since $\text{1 mol}$ of "Ba"("OH")_2 brings ${\text{2 mols OH}}^{-}$.

Here you have $\text{325 mEq/L}$ "Ba"("OH")_2, which means you have:

(325 cancel("mEq Ba"("OH")_2))/"L" xx (2 cancel"milli""mol OH"^(-))/(2 cancel("mEq Ba"("OH")_2)) xx (1 cancel"thing")/(1000 cancel"milli"cancel"things")

$= {\text{0.325 M OH}}^{-}$ in solution

Exact neutralization requires the same number of $\text{mols}$ as specified in the reaction, so then we have:

${\text{Ba"("OH")_2(aq) + 2"HCl"(aq) -> 2"H"_2"O"(l) + "BaCl}}_{2} \left(a q\right)$

${\text{0.325 mol OH"^(-)/cancel"L" xx 0.215 cancel"L" = "0.0699 mols OH}}^{-}$

Based on the balanced chemical reaction, $\text{2 mol HCl}$ are needed to react with "1 mol Ba"("OH")_2... but since we have this in terms of ${\text{OH}}^{-}$ already, ${\text{0.0699 mols H}}^{+}$ from $\text{HCl}$ react.

This number of mols can be contained in any volume of solution, but we specify that the concentration available is:

"1.25 % by mass" = ("1.25 g HCl")/("100 g solution")

(the solution volume is exact, being a definition of percent, so it has infinite sig figs.)

Using the density of the solution, we can convert this molarities, since we have $\left[{\text{OH}}^{-}\right]$ in molarities already.

I would treat the numerator and denominator separately:

NUMERATOR

1.25 cancel"g HCl" xx ("1 mol HCl")/(36.461 cancel"g HCl")

$=$ $\text{0.0343 mols HCl}$ (in $\text{100 g solution}$)

DENOMINATOR

100 cancel"g solution"xx overbrace(cancel"1 mL solution"/(1.08 cancel"g solution"))^("density = 1.08 g/mL") xx "1 L"/(1000 cancel"mL")

$=$ $\text{0.0926 L solution}$

Therefore, the solution molarity is:

["HCl"] = "0.0343 mols HCl"/"0.0926 L solution" = "0.370 M"

Finally, we can use this concentration to see how much volume is needed to contain ${\text{0.0699 mols H}}^{+}$:

color(blue)(V_(HCl(aq))) = 0.0699 cancel"mols HCl" xx "L"/(0.370 cancel"mol HCl")

$=$ $\text{0.189 L}$

$=$ $\textcolor{b l u e}{\text{189 mL}}$

This should make sense, because the molar concentration of ${\text{OH}}^{-}$ is less than that of ${\text{H}}^{+}$ ($\text{0.325 M}$ vs. $\text{0.370 M}$), so the volume of "Ba"("OH")_2 should be larger, not smaller.

So, it's good that we got LESS than $\text{215 mL}$, actually.

Jul 4, 2018

#### Explanation:

$\frac{1 \setminus \textrm{m L} H C l \left(s o \ln\right)}{1.08 \setminus \textrm{g} H C l \left(s o \ln\right)} \times \frac{100 \setminus \textrm{g} H C l \left(s o \ln\right)}{1.25 \setminus \textrm{g} H C l} \times \frac{36.46 \setminus \textrm{g} H C l}{\setminus \textrm{m o l} H C l} \times \frac{1 \setminus \textrm{E q} B a {\left(O H\right)}_{2}}{1000 \setminus \textrm{m E q} B a {\left(O H\right)}_{2}} \times \frac{325 \setminus \textrm{m E q} B a {\left(O H\right)}_{2}}{\setminus \textrm{L} B a {\left(O H\right)}_{2} \left(s o \ln\right)} \times \frac{1 \setminus \textrm{L} B a {\left(O H\right)}_{2} \left(s o \ln\right)}{1000 \setminus \textrm{m L} B a {\left(O H\right)}_{2} \left(s o \ln\right)} \times \frac{215 \setminus \textrm{m l} B a {\left(O H\right)}_{2} \left(s o \ln\right)}{1}$

$= 188.9 \setminus \textrm{m L} H C l \left(s o \ln\right) \setminus \Rightarrow 189 \setminus \textrm{m L} H C l \left(s o \ln\right)$