Question

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.

Answer 1

The volume needed is `"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.

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Equivalents are defined with respect to the `"OH"^(-)` or `"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^@ "C"` is `"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 `"1 mol"` of `"Ba"("OH")_2` brings `"2 mols OH"^(-)`.

Here you have `"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")`

`= "0.325 M OH"^(-)` in solution

Exact neutralization requires the same number of `"mols"` as specified in the reaction, so then we have:

`"Ba"("OH")_2(aq) + 2"HCl"(aq) -> 2"H"_2"O"(l) + "BaCl"_2(aq)`

`"0.325 mol OH"^(-)/cancel"L" xx 0.215 cancel"L" = "0.0699 mols OH"^(-)`

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

`=` `"0.0343 mols HCl"` (in `"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")`

`=` `"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 `"0.0699 mols H"^(+)`:

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

`=` `"0.189 L"`

`=` `color(blue)("189 mL")`

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

So, it's good that we got LESS than `"215 mL"`, actually.

Answer 2

computer webcam

Explanation:

`(1\text( mL )HCl(soln))/(1.08\text( g )HCl(soln))xx(100\text( g )HCl(soln))/(1.25\text( g )HCl)xx(36.46\text( g )HCl)/(\text(mol )HCl)xx(1\text( Eq )Ba(OH)_2)/(1000\text( mEq )Ba(OH)_2)xx(325\text( mEq )Ba(OH)_2)/(\text(L )Ba(OH)_2(soln))xx(1\text( L )Ba(OH)_2(soln))/(1000\text( mL )Ba(OH)_2(soln))xx(215\text( ml )Ba(OH)_2(soln))/1`

`=188.9\text( mL )HCl(soln)\rArr189\text( mL )HCl(soln)`