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?

1 Answer
Meave60
May 12, 2018

The concentration of sulfuric acid in the initial sample is #"6.090 M"#.

Explanation:

Balanced equation

#"H"_2"SO"_4("aq") + "2NaOH(aq)"##rarr##"Na"_2"SO"_4("aq") + "2H"_2"O("l")"#

Titration formula:

#M_"acid"V_"acid"=M_"base"V_"base"#

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

Known

#V_"acid"=12.23"mL"xx"1 L"/"1000 mL"="0.01223 L"#

#M_"base"="1.013 M"="1.013 mol/L"#

#V_"base"=36.76"mL"xx"1 L"/"1000 mL"="0.03676 L"#

Unknown

#M_"acid"

Solution

Rearrange the equation to isolate #M_"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_"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 #"6.090 M"#.

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