Question

A 10 mL sample of hydrochloric acid neutralizes 15 mL of a 0.40 M solution of `NaOH`. What is the molarity of the hydrochloric acid?

Answer 1

`"0.60 M"`

Explanation:

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

As you can see, it takes one mole of hydrochloric acid to neutralize one mole of strong base. This means that a complete neutralization requires equal numbers of moles of the two reactants.

Even without doing any calculations, you can look at the values given to you and try to estimate the concentration of the hydrochloric acid solution relative to that of the sodium hydroxide.

You know that `"10 mL"` of this hydrochloric acid solution must contain the same number of moles of solute as `"15 mL"` of sodium hydroxide solution.

Since you have the same number of moles in a smaller volume, it follows that the hydrochloric acid solution will be more concentrated than the sodium hydroxide solution.

`color(green)(|bar(ul(color(white)(a/a)color(black)(["HCl"] > "0.40 M")color(white)(a/a)|)))`

So, use the molarity and volume of the sodium hydroxide solution to determine how many moles of solute it contains

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))`

You will find

`n_(NaOH) = "0.40 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))`

`n_(NaOH) = "0.00600 moles NaOH"`

Now simply use the volume of the hydrochloric acid solution to find its concentration

`["HCl"] = "0.00600 moles"/(10 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.60 M"color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs.

As predicted, the hydrochloric acid solution was more concentrated than the sodium hydroxide solution.

`["HCl"] > ["NaOH"] color(white)(a)color(green)(sqrt())`