What volume of 33% #"HCl"# required to neutralize 8000 gal of 50 % #"NaOH"#?

For purposes of calculation, I make several assumptions:

1. Your percentages are mass percent.
2. You are using Imperial gallons.
3. The density of a 33 % #"HCl"# solution is 1.164 kg/L.
4. The density of a 50 % #"NaOH"# solution is 1.525 kg/L

Step 1: Use the balanced equation to get the mass ratios.

#color(white)(X)"HCl" color(white)(X)+ color(white)(X)"NaOH" → "NaCl" + "H"_2"O"#
#"36.46 kg" color(white)(XX)"40.00 kg"#

So, you need 36.46 kg of #"HCl"# to neutralize 40.00 kg of #"NaOH"#.

Step 2: Calculate the mass of #"NaOH"#.

#8000 color(red)(cancel(color(black)("gal soln"))) × (4.546 color(red)(cancel(color(black)("L soln"))))/(1 color(red)(cancel(color(black)("gal soln")))) × (1.525 color(red)(cancel(color(black)("kg soln"))))/(1 color(red)(cancel(color(black)("L soln")))) × "50 kg NaOH"/(100 color(red)(cancel(color(black)("kg soln")))) = "27 731 kg NaOH"#

Step 3: Calculate the mass of #"HCl"# required.

#"27 731" color(red)(cancel(color(black)("kg NaOH"))) × "36.46 kg HCl"/(40.00 color(red)(cancel(color(black)("kg NaOH")))) = "25 276 kg HCl"#

Step 4: Calculate the volume of #"HCl"# solution required.

#"25 276" color(red)(cancel(color(black)("kg HCl"))) × (100 color(red)(cancel(color(black)("kg soln"))))/(33 color(red)(cancel(color(black)("kg HCl")))) × (1 color(red)(cancel(color(black)("L soln"))))/(1.164 color(red)(cancel(color(black)("kg soln")))) × "1 gal soln"/(4.546 color(red)(cancel(color(black)("L soln")))) = "14 500 gal soln"#

You will need to use 14 500 gal of 33 % #"HCl"#.

Let's solve this problem as if it were a titration calculation.

We will need the molarities of the #"NaOH"# and #"HCl"# solutions.

Molarity of #bb"NaOH"#

Assume that we have 1 L of the #"NaOH"# solution.

#"Mass of NaOH" = "1000" color(red)(cancel(color(black)("mL soln"))) × (1.525 color(red)(cancel(color(black)("g soln"))))/(1 color(red)(cancel(color(black)("mL soln")))) × "50 g NaOH"/(100 color(red)(cancel(color(black)("g soln")))) = "762.5 g NaOH"#

#"Moles of NaOH" = "762.5"color(red)(cancel(color(black)("g NaOH"))) ×"1 mol NaOH"/(40.00 color(red)(cancel(color(black)("g NaOH")))) = "19.06 mol NaOH"#

#"Molarity" = "moles"/"litres" = "19.06 mol"/"1 L" = "19.06 mol/L"#

Molarity of #bb"HCl"#

Assume that you have 1 L of #"HCl"# solution.

#"Mass of HCl" = 1000 color(red)(cancel(color(black)("mL soln"))) × (1.164 color(red)(cancel(color(black)("g soln"))))/(1 color(red)(cancel(color(black)("mL soln")))) × "33 g HCl"/(100 color(red)(cancel(color(black)("g soln")))) = "384 g HCl"#

#"Moles of HCl" = 384 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "10.54 mol HCl"#

#"Molarity" = "moles"/"litres" = "10.54 mol"/"1 L" = "10.54 mol/L"#

Titration Calculation

The question is now, "What volume of 10.54 mol/L #"HCl"# is required to neutralize 8000 gal of 19.06 mol/L #"NaOH"#?"

Since 1 mol #"HCl"# reacts with 1 mol #"NaOH"#, we can use the formula

#c_aV_a = c_bV_b#

or

#V_a = V_b × c_b/c_a = "8000 gal" × (19.06 color(red)(cancel(color(black)("mol/L"))))/(10.54 color(red)(cancel(color(black)("mol/L")))) = "14 500 gal"#

The volume of #"HCl"# is 80 % greater than the volume of #"NaOH"#, because the #"NaOH"# is 80 % more concentrated than the #"HCl"#.