# Question b372f

Jun 15, 2016

$\text{189 kg}$

#### Explanation:

You're looking to make a $\text{25% m/v}$ solution, that is, a solution that has a 25% mass by volume percent concentration.

A solution's mass by volume percent concentration tells you the mass of solute, usually expressed in grams, you get per $\text{100 mL}$ of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% m/v" = "grams of solute"/"100 mL of solution} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, a $\text{25% m/v}$ sodium hydroxide solution will contain $\text{25 g}$ of sodium hydroxide for every $\text{100 mL}$ of solution. This is equivalent to saying that every milliliter of solution will contain

1 color(red)(cancel(color(black)("mL solution"))) * overbrace("25 g NaOH"/(100color(red)(cancel(color(black)("mL solution")))))^(color(darkgreen)("= 25% m/v")) = "0.25 g NaOH"

So, you know that your solution must contain $\text{0.25 g}$ of sodium hydroxide per milliliter of solution. All you have to do now is figure out how many milliliters you get in $200$ gallons by using the conversion factors

color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(a/a)|)))" " and " "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 gal " = " 3.785 L")color(white)(a/a)|)))

You will get

200 color(red)(cancel(color(black)("gal"))) * (3.785color(red)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("gal")))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "757,000 mL"

Since $\text{1 mL}$ must contain $\text{0.25 g}$ of sodium hydroxide, it follows that $\text{757,000 mL}$ must contain

$\text{757,000" color(red)(cancel(color(black)("mL solution"))) * "0.25 g NaOH"/(1color(red)(cancel(color(black)("mL solution")))) = "189,250 g NaOH}$

I'll leave this rounded off to three sig figs and expressed in kilograms

"mass of NaOH" = color(green)(|bar(ul(color(white)(a/a)color(black)("189 kg")color(white)(a/a)|)))#

Now, it's absolutely crucial to remember that you must never add water to solid sodium hydroxide! The reaction is highly exothermic, i.e. it releases considerable amounts of heat, and can cause some of the solution to splatter.

Keep in mind that sodium hydroxide solutions are highly corrosive, so never try to handle them without proper protection.

You should always add small quantities of sodium hydroxide to a large volume of water until you dissolve all the solid that is needed for the target solution.

Once this is done, dilute the resulting solution to the final volume of the target solution.