Question

Question #aa2ed

Answer 1

`"0.9 M"`

Explanation:

Start by writing a balanced chemical equation that describes this neutralization reaction.

`"H"_ 2"SO"_ (4(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + "H"_ 2"O"_ ((l))`

Now, notice that it takes `2` moles of sodium hydroxide to neutralize `1` mole of sulfuric acid. This means that in order for a complete neutralization to take place, the solution must contain twice as many moles of sodium hydroxide than of sulfuric acid.

`color(blue)(ul(color(black)(2 xx "moles of H"_2"SO"_4 = "moles of NaOH")))`

In your case, you know that a solution of sulfuric acid that has a molarity of `"3 M"` and a volume of `"11.56 mL"` was required to neutralize a solution of sodium hydroxide that has a volume of `"25.00 mL"`.

As you know, the number of moles of sulfuric acid needed can be calculated by using the molarity and the volume of the solution.

`11.56 color(red)(cancel(color(black)("mL solution"))) * ("3 moles H"_2"SO"_4)/(10^3color(red)(cancel(color(black)("mL solution")))) = ( (11.56 * 3)/10^3)color(white)(.)"moles H"_2"SO"_4`

If you take `x` `"M"` to be the molarity of the sodium hydroxide solution

`"molarity NaOH" = x color(white)(.)"M"`

you can say that the neutralization required

`25.00 color(red)(cancel(color(black)("mL solution"))) * (xcolor(white)(.)"moles NaOH")/(10^3color(red)(cancel(color(black)("mL solution")))) = ((25.00 * x)/10^3)color(white)(.)"moles NaOH"`

Now all you have to do is use the aforementioned `1:2` mole ratio that exists between the two reactants to say that the number of moles of sodium hydroxide must be two times the number of moles of sulfuric acid.

`2 " " xx " " overbrace((11.56 * 3)/color(red)(cancel(color(black)(10^3))))^(color(blue)("no. of moles of H"_2"SO"_4)) = overbrace((25.00 * x)/color(red)(cancel(color(black)(10^3))))^(color(blue)("no. of moles of NaOH"))`

Now just solve for `x` to find

`2 * 11.56 * 3 = 25.00 * x`

`x = (2 * 11.56 * 3)/25.00 = 0.9`

This means that the molarity of the sodium hydroxide solution is equal to

`color(darkgreen)(ul(color(black)("molarity NaOH = 0.9 M")))`

The answer is rounded to one significant figure, the number of sig figs you have for the molarity of the sulfuric acid solution.