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 "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 "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.

Jul 4, 2018

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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)