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cinderalla16dh
15.07.2019 •
Chemistry
What volume of a 0.3300-m solution of sodium hydroxide would be required to titrate 15.00 ml of 0.1500 m oxalic acid? c2 o4 h2(aq) + 2naoh(aq) ⟶ na2 c2 o4(aq) + 2h2 o(l)?
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Ответ:
13.64 ml
Explanation:
1) Titration will be done when all the moles of the oxalic acid have reacted with the volume of sodium hydroxide.
2) The balanced chemical equation is:
C₂O₄H₂ (aq) + 2 NaOH(aq) ⟶ Na₂C₂O₄(aq) + 2H₂O(l)
3) Mole ratio:
1 mol C₂O₄H₂ (aq) ; 2 mole NaOH(aq)
4) Calculate the number of moles of C₂O₄H₂ (aq)
M = n/V ⇒ n = MV = 0.01500 l × 0.1500M / 1000 = 0.00225 mol
5) Calculate the number of moles of NaOH needed using proportion with the mole ratio:
1 mol C₂O₄H₂ (aq) / 2 mole NaOH(aq) = 0.00225 mol C₂O₄H₂ (aq) / x
⇒ x = 0.0045 mol NaOH
6) Use molarity to find the volume that contains 0.0045 mol of NaOH.
M = n / V ⇒ V = n / M = 0.0045 / 0.3300 = 0.1363 liter = 13.64 ml
Ответ: