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You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeyer flask and add 12.2 mL of distilled water and a drop of phenopthalein to the flask. You fill your buret with 0.103 M NaOH (aq) solution and begin the titration. During the titration you rinse the tip and the sides of the Erlenmeyer flask with 3.52 mL of distilled water. It requires 10.38 mL of your NaOH (aq) solution to reach the endpoint and a very faint pink color in the flask. What is the concentration of your sulfuric acid solution in M (Remember the balanced equation for this reaction)?

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Respuesta :

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

[tex]V[/tex] = 10 mL solution

[tex]V_{added}[/tex] = 12.20 mL

[tex]V_{total}[/tex] = 22.20 mL

then,

M 0.103 M of NaOH,

[tex]V_{rinsed}[/tex] = experiment  will not be affected

[tex]V_{total \ base}[/tex] = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = [tex]0.103\times 10.38[/tex]

                            =  [tex]1.06914 \ m[/tex]

Whether Sulfuric acid, then

⇒  [tex]H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O[/tex]

⇒  [tex]mol \ of \ acid =\frac{1}{2}\times \ mol \ of \ base[/tex]

⇒  [tex]1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid[/tex]

Before any dilution:

[tex]V_{sample} = 10 \ mL[/tex]

⇒  [tex]M \ acid = \frac{m \ mol}{V}[/tex]

                 [tex]=\frac{ 0.53457 }{10}[/tex]

                 [tex]=0.053457 \ M[/tex] (Sulfuric acid)