Question

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"g"/"mL"`.

`2"HCl (aq)"+"Ba(OH)"_2"(aq)"\to2"H"_2"O (l)"+"BaCl"_2"(aq)"`

Answer 1

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 `"93.8 mL"` of `"0.660 M HCl"` reacts with `"375 mL"` of `"0.0825 M Ba"("OH")_2`.

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As before, a milliequivalent is just the number of millimols with respect to `"OH"^(-)` for a strong base or `"H"^(+)` for a strong acid. All we have different here are the numbers.

What we do is convert the `"mEq/L"` to `"mol/L"` of `"OH"^(-)`, and convert the `"%w/w"` of `"HCl"` to `"mol/L"` as well.

BASE CONCENTRATION

As mentioned, `"1 Eq" = "1 mol OH"^(-)` in strong bases or `"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))`

`= ul("0.165 M OH"^(-))`

(NOTE: since there are approximately two `"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, `"1 mmol"/"mL"` is the same as `"1 milli"/"1 milli" xx "1 mol"/"L"`.

ACID CONCENTRATION

Percent by mass, or `"%w/w"`, is the mass of solute in `"100 g"` of the system:

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

So, use the molar mass of `"HCl"` to get to `"mols"`, and use the density of the solution to get to `"L"`, thereby giving you `"mol/L"`. Since `"HCl"` is a strong acid:

`["H"^(+)] = ["HCl"]`

and so, treating the numerator and denominator separately:

`["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")`

`= ul("0.660 M H"^(+))`

To evaluate this, just evaluate the top and bottom separately and then you should get `"0.0590 mols"` of `"HCl"` dissolved in `"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 `"H"^(+)` and `"OH"^(-)`, we just react them exactly.

We know the `"OH"^(-)` is contained in `"375 mL"`, so the mols are:

`0.375 cancel"L" xx "0.165 mol"/cancel"L" = "0.0619 mols OH"^(-)`

And we know we want exact reaction, so we want `"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")`