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Consider the following balanced equation:H₂SO4 (aq) + 2LiOH (aq) --> 2 H₂O (1) + Li₂SO4 (aq)A 33.98 mL sample of H₂SO4 is completely titrated with 18.19 mL of 1.35 M LIOH. What is the molarity of the H₂SO4?

Sagot :

1) Write the chemical equation.

[tex]H_2SO_{4(aq)}+2LiOH\rightarrow2H_2O+Li_2SO_4[/tex]

2) List the known and unknown quantities.

Sample: H2SO4

Volume: 33.98 mL.

Concentration: unknown

Titrant: LiOH

Volume: 18.19 mL = 0.01819

Concentration: 1.35 M

3) Find moles of LiOH

3.1- Set the equation

[tex]M=\frac{moles\text{ }of\text{ }solute}{liters\text{ }of\text{ }solution}[/tex]

3.2- Plug in the values

[tex]1.35\text{ }M=\frac{moles\text{ }of\text{ }solute}{0.01819\text{ }L}[/tex][tex]moles\text{ }of\text{ }solute=1.35\text{ }M*0.01819\text{ }L[/tex][tex]moles\text{ }LiOH=0.0245565[/tex]

The number of moles of LiOH in the reaction is 0.0246.

4) Find moles of H2SO4

The molar ratio between H2SO4 and LiOH is 1 mol H2SO4: 2 mol LiOH.

[tex]mol\text{ }H_2SO_4=0.0246\text{ }mol\text{ }LiOH*\frac{1\text{ }mol\text{ }H_2SO_4}{2\text{ }mol\text{ }LiOH}[/tex][tex]mol\text{ }H_2SO_4=0.01228\text{ }mol\text{ }H_2SO_4[/tex]

5) Molarity of H2SO4

Sample: H2SO4

Moles: 0.01228

Volume: 33.98 mL.

5.1- Convert mL to L.

1 L = 1000 mL

[tex]L=33.98\text{ }mL*\frac{1\text{ }L}{1000\text{ }mL}=0.03398\text{ }L[/tex]

5.2- Set the equation to find molarity.

[tex]M=\frac{moles\text{ }of\text{ }solute}{liters\text{ }of\text{ }solution}[/tex][tex]M=\frac{0.01228\text{ }mol\text{ }H_2SO_4}{0.03398\text{ }L}[/tex][tex]M=0.3614\text{ }M[/tex]

The molarity of the sample is 0.3614 M H2SO4.

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