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)
