# Question d0e46

Feb 21, 2017

$\text{0.375 M}$

#### Explanation:

Sodium hydroxide and hydrobromic acid will react in a $1 : 1$ mole ratio to produce aqueous sodium bromide and water

${\text{NaOH"_ ((aq)) + "HBr"_ ((aq)) -> "NaBr"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

As you can see, this neutralization reaction consumes equal numbers of moles of strong acid and of strong base.

Use the molarity of the hydrobromic acid solution to calculate how many moles were needed to neutralize the sodium hydroxide

30.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.250 moles HBr"/(1 color(red)(cancel(color(black)("L solution")))) = "0.00750 moles HBr"

This, of course, tells you that the sample of sodium hydroxide contained $0.00750$ moles of solute. To find the molarity of the sodium hydroxide solution, you must find the number of moles of solute present in $1$ liter of solution

1 color(red)(cancel(color(black)("L solution"))) * (10^3 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.00750 moles NaOH"/(20.0 color(red)(cancel(color(black)("mL")))) = "0.375 moles NaOH"#

Since this represents the number of moles of sodium hydroxide present in $\text{1 L}$ of solution, you can say that its molarity will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity NaOH = 0.375 mol L}}^{- 1}}}}$

The answer is rounded to three sig figs.