# A student observes that 36.76 mL of 1.013 M NaOH are required to neutralize a 12.23 mL aqueous solution of sulfuric acid. What is the concentration of sulfuric acid in the initial sample?

May 12, 2018

The concentration of sulfuric acid in the initial sample is $\text{6.090 M}$.

#### Explanation:

Balanced equation

$\text{H"_2"SO"_4("aq") + "2NaOH(aq)}$$\rightarrow$$\text{Na"_2"SO"_4("aq") + "2H"_2"O("l")}$

Titration formula:

${M}_{\text{acid"V_"acid"=M_"base"V_"base}}$

Since molarity is mol/L, you will need to convert mL to L.

Known

${V}_{\text{acid"=12.23"mL"xx"1 L"/"1000 mL"="0.01223 L}}$

${M}_{\text{base"="1.013 M"="1.013 mol/L}}$

${V}_{\text{base"=36.76"mL"xx"1 L"/"1000 mL"="0.03676 L}}$

Unknown

M_"acid"

Solution

Rearrange the equation to isolate ${M}_{\text{acid}}$. Plug in known values and solve. The molarity of the base will be multiplied by $2$ because the mole ratio between the molarity of the base and the molarity of the acid is $2 : 1$.

M_"acid"=(2xxM_"base"xxV_"base")/(V_"acid")#

${M}_{\text{acid"=(2xx1.013"mol"/color(red)cancel(color(black)("L"))xx0.03676color(red)cancel(color(black)("L")))/(0.01223"L")="6.090 mol/L"="6.090 M}}$ (rounded to four significant figures)

The concentration of sulfuric acid in the initial sample is $\text{6.090 M}$.