Question

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

Answer 1

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

Explanation:

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"`.

Answer 2

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

Explanation:

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"`.