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

Question

Natalie Anton · Accepted Answer

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

I make a few assumptions in order to calculate: Step 1: To find the mass ratios, use the balanced equation. #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"#.

Ellie Andrews · Answer

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

Assuming that this is a titration calculation, let's tackle its solution. We will need the molarities of the #"NaOH"# and #"HCl"# solutions. The molarity of "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"# The molarity of "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 Estimation 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"#.

Cooper Anderson · Answer

undefined To find the volume of 33% HCl needed to neutralize 8000 gallons of 50% NaOH, you can use the equation:

$$ \text{Volume}_{\text{HCl}} \times \text{Concentration}_{\text{HCl}} = \text{Volume}_{\text{NaOH}} \times \text{Concentration}_{\text{NaOH}} $$

$$ \text{Volume}_{\text{HCl}} \times 0.33 = 8000 \times 0.50 $$

$$ \text{Volume}_{\text{HCl}} = \frac{8000 \times 0.50}{0.33} $$

$$ \text{Volume}_{\text{HCl}} \approx 12,121.21 \, \text{gallons} $$