How many #"mL"# of #2.15%# (by mass) #"HCl"# would be required to neutralize #"375 mL"# of #"165 mEq/L"# #"Ba(OH)"_2#?

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.

BASE CONCENTRATION

Here we simply treated "milli" as a separate object.

ACID CONCENTRATION

and so, treating the numerator and denominator separately:

ACID/BASE REACTION

To determine the volume of 2.15% HCl required to neutralize 375 mL of 165 mEq/L Ba(OH)2, first calculate the number of equivalents (Eq) of Ba(OH)2:

[375 \text{ mL} \times 165 \text{ mEq/L} = 61875 \text{ mEq}]

Since Ba(OH)2 is a strong base, it will produce two equivalents of OH^- ions for every mole of Ba(OH)2:

[61875 \text{ mEq} \times 2 = 123750 \text{ mEq OH}^-]

Since HCl is a strong acid, it will neutralize OH^- ions in a 1:1 ratio:

[123750 \text{ mEq OH}^- = 123750 \text{ mEq H}^+]

Now, calculate the volume of 2.15% HCl (by mass) required to obtain 123750 mEq of H+ ions:

[123750 \text{ mEq} \div \text{equivalent weight of HCl}]

The equivalent weight of HCl is the molecular weight divided by the number of replaceable hydrogen ions (1 for HCl):

[36.46 \text{ g/mol} \div 1 = 36.46 \text{ g/mol}]

Now, calculate the mass of HCl needed:

[123750 \text{ mEq} \times 36.46 \text{ g/mol} \times 0.0215 = \text{mass of HCl in grams}]

Finally, convert the mass of HCl to milliliters using the density of 2.15% HCl (which is close to the density of water):

[\text{mass of HCl in grams} \div \text{density of 2.15% HCl} \approx \text{volume of 2.15% HCl in mL}]