Question

Question #47d01

Answer 1

`"0.50 mol L"^(-1)`

Explanation:

The first thing to do here is write a balanced chemical equation for this neutralization reaction.

Hhdrochloric acid, `"HCl"`, a strong acid, will react with sodium hydroxide, `"NaOH"`, a strong base, to produce water and aqueous sodium chloride, `"NaCl"`.

`"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaCl"_ ((aq))`

Notice that the two reactants are reacting in a `1:1` mole ratio, which means that the reaction will always consume equal number of moles of hydrochloric acid and sodium hydroxide.

In order to completely neutralize the initial number of moles of hydrochloric acid, you will need to add the same number of moles of sodium hydroxide.

Your goal here will be to determine how many moles of sodium hydroxide will neutralize your acid, then figure out what concentration of sodium hydroxide solution will contain that many moles in `"40. mL"`.

As you know, a solution's molarity tells you the number of moles of solute in one liter of solution. The hydrochloric acid solution is said to have a molarity of `"1.0 mol L"^(-1)`, which means that one liter of this solution will contain one mole of acid.

Convert the volume of the solution from milliliters to liters by using the conversion factor

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

In your case, you will have

`20. color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3 color(red)(cancel(color(black)("mL")))) = 20. * 10^(-3)"L"`

So, if the hydrochloric acid solution contains one mole of acid in one liter of solution, you will have

`20. * 10^(-3)color(red)(cancel(color(black)("L"))) * overbrace("1 mole HCl"/(1color(red)(cancel(color(black)("L")))))^(color(purple)("equivalent to 1.0 M")) = "0.020 moles HCl"`

This is exactly how many moles of sodium hydroxide you must have in the sodium hydroxide solution.

Well, if `"40. mL"` must contain `0.20` moles, it follows that one liter will contain

`1.0color(red)(cancel(color(black)("L"))) * "0.20 moles NaOH"/(40. * 10^(-3)color(red)(cancel(color(black)("L")))) = "0.50 moles NaOH"`

Therefore, the molarity of the sodium hydroxide solution is

`c_(NaOH) = color(green)(|bar(ul(color(white)(a/a)"0.50 mol L"^(-1) = "0.50 M"color(white)(a/a)|)))`

The answer is rounded to two sig figs.