An Erlenmeyer flask containing 20.0 mL of sulfuric acid of an unknown concentration was titrated with

3 answers
Question:

An Erlenmeyer flask containing 20.0 mL of sulfuric acid of an unknown concentration was titrated with exactly 15.0 mL of 0.25 M NaOH solution to the second equivalence point. What was the concentration of sulfuric acid in the flask? H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)

Answers

The concentration of sulfuric acid in the flask is  [tex]M_{H_2SO_4}= 0.09375M[/tex]

Explanation:

The number of moles is mathematically defined as

                     [tex]n = concentration * volume[/tex]

  For the number of moles of NaOH substituting 15.0mL = [tex]15 *10^{-3} L[/tex] for volume and 0.25M for concentration we have

                        [tex]n_{NaOH} = 15 *10^{-3} * 0.25[/tex]

                           [tex]= 0.00375\ moles[/tex]

From the balanced equation we can that the number of moles of [tex]H_2S0_4[/tex] required for this reaction is

                   [tex]n_{H_2SO_4} = \frac{1}{2} * n_{NaOH}[/tex]

                          [tex]= 0.5 *0.00375[/tex]

                          [tex]= 0.001875 \ moles[/tex]

Concentration is mathematically defined as            

               [tex]concentration = \frac{n}{volume}[/tex]

Concentration of [tex]H_2SO_4[/tex]  

                [tex]M_{H_2SO_4} =\frac{0.001875 }{20*10^{-3}}[/tex]

                       [tex]M_{H_2SO_4}= 0.09375M[/tex]

 

the concentration of sulfuric acid in the flask is  0.375 M

Explanation:

H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)  

Moles of NaOH = 15 x 0.25 /1000

                        = 0.00375

Moles of H2SO4 needed to neutralize = 0.00375 /2

                       = 0.001875

Molarity of H2SO4

                      = 0.001875 x 1000 /20

                      = 0.09375 M

concentration of sulfuric acid in the flask 0.09375 M

Moles

V2 = 20 ml

N2 =?

Substitute in the equation and find N2

N2 = 2 x 15 x 0.25 / 20

     = 0.375 M

Thus, the concentration of sulfuric acid in the flask is  0.375 M

0.09375M

Explanation:

There are two methods of going about this, either we use dilution formula (easiest and fastest) or we solve through molarity.

Using dilution formula,

C2 (H2SO4) = ?

C1 (NaOH) = 0.25M

V2 (H2SO4) = 20cm³

V1 (NaOH) = 15cm³

However we can solve using molarity method

Equation of reaction =

2NaOH + H2SO4 》 Na2SO4 + 2H2O

O.25M of NaOH = 1000cm³

X moles = 15cm³

X = (0.25 * 15) / 1000

X = 0.00375 moles is present in 15cm³ of NaOH

From equation of reaction,

2 moles of NaOH requires 1 mole of H2SO4

Therefore

0.00375 / 2 = 0.001875 moles is present in H2SO4

From the reaction,

0.00187 moles of H2SO4 = 20 cm³

X moles = 1000cm³

X = (0.00187*1000) / 20 = 0.09375M

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