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.

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.
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Answer 2

#(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)#
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Answer 3

To determine the volume of 1.25% (by mass) HCl needed to neutralize 215 mL of 325 mEq/L Ba(OH)2, we use the equation:

[ \text{Volume}_1 \times \text{Concentration}_1 = \text{Volume}_2 \times \text{Concentration}_2 ]

[ \text{Volume}_1 \times 1.25% = 215 \text{ mL} \times 325 \text{ mEq/L} ]

[ \text{Volume}_1 = \frac{215 \times 325}{1.25%} ]

[ \text{Volume}_1 ≈ 55928 \text{ mL} ]

So, approximately 55928 mL of 1.25% HCl would be required to neutralize 215 mL of 325 mEq/L Ba(OH)2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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