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

Dec 3, 2015

You will need to use 14 500 gal of 33 % $\text{HCl}$.

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 % $\text{HCl}$ solution is 1.164 kg/L.
4. The density of a 50 % $\text{NaOH}$ solution is 1.525 kg/L

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

$\textcolor{w h i t e}{X} \text{HCl" color(white)(X)+ color(white)(X)"NaOH" → "NaCl" + "H"_2"O}$
$\text{36.46 kg" color(white)(XX)"40.00 kg}$

So, you need 36.46 kg of $\text{HCl}$ to neutralize 40.00 kg of $\text{NaOH}$.

Step 2: Calculate the mass of $\text{NaOH}$.

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 $\text{HCl}$ required.

$\text{27 731" color(red)(cancel(color(black)("kg NaOH"))) × "36.46 kg HCl"/(40.00 color(red)(cancel(color(black)("kg NaOH")))) = "25 276 kg HCl}$

Step 4: Calculate the volume of $\text{HCl}$ solution required.

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

You will need to use 14 500 gal of 33 % $\text{HCl}$.

Dec 6, 2015

You will need to use 14 500 gal of 33 % $\text{HCl}$.

Explanation:

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

We will need the molarities of the $\text{NaOH}$ and $\text{HCl}$ solutions.

Molarity of $\boldsymbol{\text{NaOH}}$

Assume that we have 1 L of the $\text{NaOH}$ solution.

$\text{Mass of NaOH" = "1000" color(red)(cancel(color(black)("mL soln"))) × (1.525 color(red)(cancel(color(black)("g soln"))))/(1 color(red)(cancel(color(black)("mL soln")))) × "50 g NaOH"/(100 color(red)(cancel(color(black)("g soln")))) = "762.5 g NaOH}$

$\text{Moles of NaOH" = "762.5"color(red)(cancel(color(black)("g NaOH"))) ×"1 mol NaOH"/(40.00 color(red)(cancel(color(black)("g NaOH")))) = "19.06 mol NaOH}$

$\text{Molarity" = "moles"/"litres" = "19.06 mol"/"1 L" = "19.06 mol/L}$

Molarity of $\boldsymbol{\text{HCl}}$

Assume that you have 1 L of $\text{HCl}$ solution.

$\text{Mass of HCl" = 1000 color(red)(cancel(color(black)("mL soln"))) × (1.164 color(red)(cancel(color(black)("g soln"))))/(1 color(red)(cancel(color(black)("mL soln")))) × "33 g HCl"/(100 color(red)(cancel(color(black)("g soln")))) = "384 g HCl}$

$\text{Moles of HCl" = 384 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "10.54 mol HCl}$

$\text{Molarity" = "moles"/"litres" = "10.54 mol"/"1 L" = "10.54 mol/L}$

Titration Calculation

The question is now, "What volume of 10.54 mol/L $\text{HCl}$ is required to neutralize 8000 gal of 19.06 mol/L $\text{NaOH}$?"

Since 1 mol $\text{HCl}$ reacts with 1 mol $\text{NaOH}$, we can use the formula

${c}_{a} {V}_{a} = {c}_{b} {V}_{b}$

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 $\text{HCl}$ is 80 % greater than the volume of $\text{NaOH}$, because the $\text{NaOH}$ is 80 % more concentrated than the $\text{HCl}$.